#set document(title: "4.1 Exponential Growth and Decay", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 4.1#h(0.6em)Exponential Growth and Decay === Exponential Growth The functions in Population Growth describe #strong[exponential growth]. During each time interval of a fixed length, the population is multiplied by a certain constant amount. In Part A, the bacteria population grows by a factor of #math.equation(block: false, alt: "3")[$3$] every day. #figure(figph[table of bacteria population], alt: "table of bacteria population", caption: none) For this reason, we say that #math.equation(block: false, alt: "3")[$3$] is the #strong[growth factor] for the function. Functions that describe exponential growth can be expressed in a standard form. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Exponential Growth] #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals P sub 0 b to the power t")[$bold(italic(P)) ( bold(italic(t)) ) = bold(italic(P))_(0) bold(italic(b))^(bold(italic(t)))$] #math.equation(block: true, alt: "where, P sub 0 equals P open parenthesis 0 close parenthesis, is the initial value ,; and, b, is the growth factor .")[$& "where " & & P_(0) = P ( 0 ) & & "is the " " initial value" , \ & "and" & & b & & "is the " " growth factor" .$] ] For the bacteria population, we have #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 100 times 3 to the power t")[$P ( t ) = 100 ⋅ 3^(t)$] so #math.equation(block: false, alt: "P sub 0 equals 100")[$P_(0) = 100$] and #math.equation(block: false, alt: "b equals 3")[$b = 3$]. How can you tell from a table of values that a function describes exponential growth? \_\_\_\_\_ #solutionbox[ During each equal time interval, the output increases by the same factor. ] How can you tell from a table of values that a function describes exponential growth? + During each equal time interval, the output increases by the same amount. + During each equal time interval, the output increases by the same factor. + During each equal time interval, the output increases as #math.equation(block: false, alt: "t to the power n")[$t^(n)$]. + During each equal time interval, the output increases towards an asymptote. #examplebox("Example 1")[][ A colony of bacteria starts with #math.equation(block: false, alt: "300")[$300$] organisms and doubles every week. + Write a formula for the population of the bacteria colony after #math.equation(block: false, alt: "t")[$t$] weeks. + How many bacteria will there be after #math.equation(block: false, alt: "8")[$8$] weeks? After #math.equation(block: false, alt: "5")[$5$] days? #solutionbox[ + The initial value of the population was #math.equation(block: false, alt: "P sub 0 equals 300")[$P_(0) = 300$], and its weekly growth factor is #math.equation(block: false, alt: "b equals 2")[$b = 2$]. Thus, a formula for the population after #math.equation(block: false, alt: "t")[$t$] weeks is #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 300 times 2 to the power t")[$P ( t ) = 300 ⋅ 2^(t)$] + After #math.equation(block: false, alt: "8")[$8$] weeks, the population will be #math.equation(block: true, alt: "P open parenthesis 8 close parenthesis equals 300 times 2 to the power 8 equals 76 , 800 bacteria")[$P ( 8 ) = 300 ⋅ 2^(8) = 76 , 800 " bacteria"$] Because #math.equation(block: false, alt: "5")[$5$] days is #math.equation(block: false, alt: "the fraction 5 over 7")[$frac(5, 7)$] of a week, after #math.equation(block: false, alt: "5")[$5$] days the population will be #math.equation(block: true, alt: "P open parenthesis the fraction 5 over 7 close parenthesis equals 300 times 2 to the power 5 / 7 equals 492.2")[$P ( frac(5, 7) ) = 300 ⋅ 2^(5 / 7) = 492.2$] We cannot have a fraction of a bacterium, so we round to the nearest whole number, #math.equation(block: false, alt: "492")[$492$]. ] ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In Examplea, note that #math.equation(block: true, alt: "300 times 2 to the power 8 not equal to 600 to the power 8")[$300 ⋅ 2^(8) ≠ 600^(8)$] According to the order of operations, we compute the power #math.equation(block: false, alt: "2 to the power 8")[$2^(8)$] first, then multiply by #math.equation(block: false, alt: "300")[$300$]. ] A population of 24 fruit flies triples every month. + Write a formula for the population of fruit flies after #math.equation(block: false, alt: "t")[$t$] months. #linebreak() #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals")[$P ( t ) =$]\_\_\_\_\_ + How many fruit flies will there be after 6 months? \_\_\_\_\_ Note: Enter (large) numbers without any commas. That is, enter "10000" rather than "10,000". #linebreak() After 3 weeks? \_\_\_\_\_ (Assume that a month equals 4 weeks, and round your answer to the nearest whole number of flies.) #solutionbox[ + #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals 24 times 3 to the power t")[$P ( t ) = 24 ⋅ 3^(t)$] + 17,496; 55 ] A population of 24 fruit flies triples every month. + Write a formula for the population of fruit flies after #math.equation(block: false, alt: "t")[$t$] months. + How many fruit flies will there be after 6 months? After 3 weeks? #solutionbox[ + #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals 24 times 3 to the power t")[$P ( t ) = 24 ⋅ 3^(t)$] + After 6 months: 17,496; after 3 weeks: 55 ] Is #math.equation(block: false, alt: "5 times 4 to the power t")[$5 ⋅ 4^(t)$] equivalent to #math.equation(block: false, alt: "20 to the power t")[$20^(t)$]? \_\_\_\_\_ #solutionbox[ No, powers are performed before products. ] Is #math.equation(block: false, alt: "5 times 4 to the power t")[$" " 5 ⋅ 4^(t) " "$] equivalent to #math.equation(block: false, alt: "20 to the power t")[$" " 20^(t)$]? + Yes, we can simplify the product. + Yes, it is exponential growth. + No, powers are performed before products. + No, it should be #math.equation(block: false, alt: "9 to the power t")[$9^(t)$]. === Growth Factors In Part B of Population Growth, the rabbit population grew by a factor of #math.equation(block: false, alt: "2")[$2$] every #math.equation(block: false, alt: "3")[$3$] months. #figure(figph[table of rabbit population], alt: "table of rabbit population", caption: none) The variable #math.equation(block: false, alt: "t")[$t$] is given in months, so to write the growth formula for this population, we divide the value of #math.equation(block: false, alt: "t")[$t$] by #math.equation(block: false, alt: "3.")[$3.$] The value of #math.equation(block: false, alt: "t / 3")[$t / 3$] gives us the number of doubling periods. #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 60 times 2 to the power t / 3")[$P ( t ) = 60 ⋅ 2^(t / 3)$] Now we need some algebra to see the growth factor for the function. We use the third law of exponents to write #math.equation(block: false, alt: "2 to the power t / 3")[$2^(t / 3)$] in another form. Recall that to raise a power to a power, we multiply exponents, so #math.equation(block: true, alt: "open parenthesis 2 to the power 1 / 3 close parenthesis to the power t equals 2 to the power t open parenthesis 1 / 3 close parenthesis equals 2 to the power t / 3")[$attach(( 2^(1 / 3) ), t: t) = 2^(t ( 1 / 3 )) = 2^(t / 3)$] The growth law for the rabbit population is thus #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 60 times open parenthesis 2 to the power 1 / 3 close parenthesis to the power t")[$P ( t ) = 60 ⋅ attach(( 2^(1 / 3) ), t: t)$] The initial value of the function is #math.equation(block: false, alt: "P sub 0 equals 60")[$P_(0) = 60$], and the growth factor is #math.equation(block: false, alt: "b equals 2 to the power 1 / 3")[$b = 2^(1 / 3)$], or approximately #math.equation(block: false, alt: "1.26")[$1.26$]. The rabbit population grows by a factor of about #math.equation(block: false, alt: "1.26")[$1.26$] every month. A population triples every four years. What is its annual growth factor? \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "3 to the power the fraction 1 over 4")[$3^(frac(1, 4))$] ] A population triples every four years. What is its annual growth factor? + #math.equation(block: false, alt: "the fraction 3 over 4")[$frac(3, 4)$] + #math.equation(block: false, alt: "the cube root of 4")[$root(3, 4)$] + #math.equation(block: false, alt: "3 to the power 4")[$3^(4)$] + #math.equation(block: false, alt: "3 to the power 1 / 4")[$3^(1 / 4)$] If the units are the same, a population with a larger growth factor grows faster than one with a smaller growth factor. #examplebox("Example 2")[][ A lab technician compares the growth of #math.equation(block: false, alt: "2")[$2$] species of bacteria. She starts #math.equation(block: false, alt: "2")[$2$] colonies of #math.equation(block: false, alt: "50")[$50$] bacteria each. Species A doubles in population every #math.equation(block: false, alt: "2")[$2$] days, and species B triples every #math.equation(block: false, alt: "3")[$3$] days. Find the growth factor for each species. #solutionbox[ A function describing the growth of species A is #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 50 times 2 to the power t / 2 equals 50 times open parenthesis 2 to the power 1 / 2 close parenthesis to the power t")[$P ( t ) = 50 ⋅ 2^(t / 2) = 50 ⋅ attach(( 2^(1 / 2) ), t: t)$] so the growth factor for species A is #math.equation(block: false, alt: "2 to the power 1 / 2")[$2^(1 / 2)$], or approximately #math.equation(block: false, alt: "1.41")[$1.41$]. For species B, #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 50 times 3 to the power t / 3 equals 50 times open parenthesis 3 to the power 1 / 3 close parenthesis to the power t")[$P ( t ) = 50 ⋅ 3^(t / 3) = 50 ⋅ attach(( 3^(1 / 3) ), t: t)$] so the growth factor for species B is #math.equation(block: false, alt: "3 to the power 1 / 3")[$3^(1 / 3)$], or approximately #math.equation(block: false, alt: "1.44")[$1.44$]. Species B grows faster than species A. ] ] In 1999, analysts expected the number of Internet service providers to double in five years. + What was the annual growth factor for the number of Internet service providers? #linebreak() Answer: \_\_\_\_\_ + If there were #math.equation(block: false, alt: "5078")[$5078$] Internet service providers in April 1999, estimate the number of providers in April 2000 and in April 2001. #linebreak() April 2000: \_\_\_\_\_ #linebreak() April 2001: \_\_\_\_\_ + Write a formula for #math.equation(block: false, alt: "I open parenthesis t close parenthesis")[$I ( t )$], the number of Internet service providers #math.equation(block: false, alt: "t")[$t$] years after 1999. #linebreak() #math.equation(block: false, alt: "I open parenthesis t close parenthesis equals")[$I ( t ) =$]\_\_\_\_\_ Source: LA Times, Sept. 6, 1999 #solutionbox[ + #math.equation(block: false, alt: "2 to the power 1 / 5")[$2^(1 / 5)$] + #math.equation(block: false, alt: "5833")[$5833$] and #math.equation(block: false, alt: "6700")[$6700$] + #math.equation(block: false, alt: "I open parenthesis t close parenthesis equals 5078 times 2 to the power t / 5")[$I ( t ) = 5078 ⋅ 2^(t / 5)$] ] In 1999, analysts expected the number of Internet service providers to double in five years. Source: LA Times, Sept. 6, 1999 + What was the annual growth factor for the number of Internet service providers? + If there were #math.equation(block: false, alt: "5078")[$5078$] Internet service providers in April 1999, estimate the number of providers in April 2000 and in April 2001. + Write a formula for #math.equation(block: false, alt: "I open parenthesis t close parenthesis")[$I ( t )$], the number of Internet service providers #math.equation(block: false, alt: "t")[$t$] years after 1999. #solutionbox[ + #math.equation(block: false, alt: "2 to the power 1 / 5")[$2^(1 / 5)$] + #math.equation(block: false, alt: "5833")[$5833$] and #math.equation(block: false, alt: "6700")[$6700$] + #math.equation(block: false, alt: "I open parenthesis t close parenthesis equals 5078 times 2 to the power t / 5")[$I ( t ) = 5078 ⋅ 2^(t / 5)$] ] === Percent Increase Exponential growth occurs in other circumstances, too. For example, if the interest on a savings account is compounded annually, the amount of money in the account grows exponentially. Consider a principal of \$100 invested at 5% interest compounded annually. At the end of #math.equation(block: false, alt: "1")[$1$] year, the amount is #math.equation(block: true, alt: "Amount, equals Principal plus Interest; A, equals P plus P r; equals 100 plus 100 open parenthesis 0.05 close parenthesis equals 105")[$"Amount" & = "Principal" + "Interest" \ A & = P + P r \ & = 100 + 100 ( 0.05 ) = 105$] It will be more useful to write the formula for the amount after #math.equation(block: false, alt: "1")[$1$] year in factored form. #math.equation(block: true, alt: "A, equals P plus P r, Factor out P.; equals P open parenthesis 1 plus r close parenthesis")[$A & = P + P r & & "Factor out P." \ & = P ( 1 + r )$] With this version of the formula, the calculation for the amount at the end of #math.equation(block: false, alt: "1")[$1$] year looks like this: #math.equation(block: true, alt: "A, equals P open parenthesis 1 plus r close parenthesis; equals 100 open parenthesis 1 plus 0.05 close parenthesis; equals 100 open parenthesis 1.05 close parenthesis equals 105")[$A & = P ( 1 + r ) \ & = 100 ( 1 + 0.05 ) \ & = 100 ( 1.05 ) = 105$] The amount, \$105, becomes the new principal for the second year. To find the amount at the end of the second year, we apply the formula again, with #math.equation(block: false, alt: "P equals 105")[$P = 105$]. #math.equation(block: true, alt: "A, equals P open parenthesis 1 plus r close parenthesis; equals 105 open parenthesis 1 plus 0.05 close parenthesis; equals 105 open parenthesis 1.05 close parenthesis equals 110.25")[$A & = P ( 1 + r ) \ & = 105 ( 1 + 0.05 ) \ & = 105 ( 1.05 ) = 110.25$] Observe that to find the amount at the end of each year, we multiply the principal by a factor of #math.equation(block: false, alt: "1 plus r equals 1.05")[$1 + r = 1.05$]. Thus, we can express the amount at the end of the second year as #math.equation(block: true, alt: "A, equals open bracket 100 open parenthesis 1.05 close parenthesis close bracket open parenthesis 1.05 close parenthesis; equals 100 open parenthesis 1.05 close parenthesis squared")[$A & = [ 100 ( 1.05 ) ] ( 1.05 ) \ & = 100 ( 1.05 )^(2)$] and at the end of the third year as #math.equation(block: true, alt: "A, equals open bracket 100 open parenthesis 1.05 close parenthesis squared close bracket open parenthesis 1.05 close parenthesis; equals 100 open parenthesis 1.05 close parenthesis cubed")[$A & = [ 100 ( 1.05 )^(2) ] ( 1.05 ) \ & = 100 ( 1.05 )^(3)$] At the end of each year, we multiply the old balance by another factor of #math.equation(block: false, alt: "1.05")[$1.05$] to get the new amount. We organize our results into a table, where #math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$] represents the amount of money in the account after #math.equation(block: false, alt: "t")[$t$] years. For this example, a formula for the amount after #math.equation(block: false, alt: "t")[$t$] years is #math.equation(block: true, alt: "A open parenthesis t close parenthesis equals 100 open parenthesis 1.05 close parenthesis to the power t")[$A ( t ) = 100 ( 1.05 )^(t)$] #figure(table( columns: 3, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "P open parenthesis 1 plus r close parenthesis to the power t")[$P ( 1 + r )^(t)$]], [#math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$]]), [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "100")[$100$]], [#math.equation(block: false, alt: "100")[$100$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "100 open parenthesis 1.05 close parenthesis")[$100 ( 1.05 )$]], [#math.equation(block: false, alt: "105")[$105$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "100 open parenthesis 1.05 close parenthesis squared")[$100 ( 1.05 )^(2)$]], [#math.equation(block: false, alt: "110.25")[$110.25$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "100 open parenthesis 1.05 close parenthesis cubed")[$100 ( 1.05 )^(3)$]], [#math.equation(block: false, alt: "115.76")[$115.76$]], )) In general, for an initial investment of #math.equation(block: false, alt: "P")[$P$] dollars at an interest rate #math.equation(block: false, alt: "r")[$r$] compounded annually, we have the following formula for the amount accumulated after #math.equation(block: false, alt: "t")[$t$] years. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Compound Interest] The #strong[amount #math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$]] accumulated (principal plus interest) in an account bearing interest compounded annually is #math.equation(block: true, alt: "A open parenthesis t close parenthesis equals P open parenthesis 1 plus r close parenthesis to the power t")[$bold(italic(A)) ( bold(italic(t)) ) = bold(italic(P)) ( 1 + bold(italic(r)) )^(bold(italic(t)))$] where #math.equation(block: true, alt: "P, is the principal invested,; r, is the interest rate,; t, is the time period, in years.")[$& P & & "is the principal invested," \ & r & & "is the interest rate," \ & t & & "is the time period, in years."$] ] This function describes exponential growth with an initial value of #math.equation(block: false, alt: "P")[$P$] and a growth factor of #math.equation(block: false, alt: "b equals 1 plus r")[$b = 1 + r$]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The notion of #strong[percent increase] is often used to describe the growth factor for quantities that grow exponentially. Note carefully the distinction between the percent increase, #math.equation(block: false, alt: "r")[$r$], and the growth factor, #math.equation(block: false, alt: "b equals 1 plus r")[$b = 1 + r$]. ] A population grows by 4.5% per year. What is its annual growth factor? \_\_\_\_\_ #solutionbox[ 1.045 ] A population grows by 4.5% per year. What is its annual growth factor? + #math.equation(block: false, alt: "4.5")[$4.5$] + #math.equation(block: false, alt: "0.45")[$0.45$] + #math.equation(block: false, alt: "0.045")[$0.045$] + #math.equation(block: false, alt: "1.045")[$1.045$] #examplebox("Example 3")[][ During a period of rapid inflation, prices rose by #math.equation(block: false, alt: "12")[$12$]% over #math.equation(block: false, alt: "6")[$6$] months. At the beginning of the inflationary period, a pound of butter cost \$#math.equation(block: false, alt: "2")[$2$]. + Make a table of values showing the rise in the cost of butter over the next #math.equation(block: false, alt: "2")[$2$] years. + Write a function that gives the price of a pound of butter #math.equation(block: false, alt: "t")[$t$] years after inflation began. + How much did a pound of butter cost after #math.equation(block: false, alt: "3")[$3$] years? After #math.equation(block: false, alt: "15")[$15$] months? + Graph the function you found in part (b). #solutionbox[ + The percent increase in the price of butter is #math.equation(block: false, alt: "12")[$12$]% every #math.equation(block: false, alt: "6")[$6$] months. Therefore, the growth factor for the price of butter is #math.equation(block: false, alt: "1 plus 0.12 equals 1.12")[$1 + 0.12 = 1.12$] every half-year. If #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$] represents the price of butter after #math.equation(block: false, alt: "t")[$t$] years, then #math.equation(block: false, alt: "P open parenthesis 0 close parenthesis equals 2")[$P ( 0 ) = 2$], and every half-year we multiply the price by #math.equation(block: false, alt: "1.12")[$1.12$], as shown in the table.#figure(figph[table for inflation], alt: "table for inflation", caption: none) + Look closely at the second column of the table. After #math.equation(block: false, alt: "t")[$t$] years of inflation, the original price of \$#math.equation(block: false, alt: "2")[$2$] has been multiplied by #math.equation(block: false, alt: "2 t")[$2 t$] factors of #math.equation(block: false, alt: "1.12")[$1.12$]. Thus, #math.equation(block: true, alt: "P equals 2 open parenthesis 1.12 close parenthesis to the power 2 t")[$P = 2 ( 1.12 )^(2 t)$] + To find the price of butter at any time after inflation began, we evaluate the function at the appropriate value of #math.equation(block: false, alt: "t")[$t$]. #math.equation(block: true, alt: "P open parenthesis 3 close parenthesis, equals 2 open parenthesis 1.12 close parenthesis to the power 2 open parenthesis 3 close parenthesis; equals 2 open parenthesis 1.12 close parenthesis to the power 6 approximately equals 3.95")[$P ( 3 ) & = 2 ( 1.12 )^(2 ( 3 )) \ & = 2 ( 1.12 )^(6) ≈ 3.95$] After #math.equation(block: false, alt: "3")[$3$] years, the price was \$#math.equation(block: false, alt: "3.95")[$3.95$]. Fifteen months is #math.equation(block: false, alt: "1.25")[$1.25$] years, so we evaluate #math.equation(block: false, alt: "P open parenthesis 1.25 close parenthesis")[$P ( 1.25 )$]. #math.equation(block: true, alt: "P open parenthesis 1.25 close parenthesis, equals 2 open parenthesis 1.12 close parenthesis to the power 2 open parenthesis 1.25 close parenthesis; equals 2 open parenthesis 1.12 close parenthesis to the power 2.5 approximately equals 2.66")[$P ( 1.25 ) & = 2 ( 1.12 )^(2 ( 1.25 )) \ & = 2 ( 1.12 )^(2.5) ≈ 2.66$] After #math.equation(block: false, alt: "15")[$15$] months, the price of butter was \$#math.equation(block: false, alt: "2.66")[$2.66$]. + Evaluate the function #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 2 open parenthesis 1.12 close parenthesis to the power 2 t")[$P ( t ) = 2 ( 1.12 )^(2 t)$] for several values, as shown in the table below. We plot the points and connect them with a smooth curve to obtain the graph shown in the figure at right below.#figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$]]), [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "2.00")[$2.00$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2.51")[$2.51$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3.15")[$3.15$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "3.95")[$3.95$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "4.95")[$4.95$]], )) #figure(figph[graph of inflation], alt: "graph of inflation", caption: none) ] ] In Example, we can rewrite the formula for #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$] as follows: #math.equation(block: true, alt: "P open parenthesis t close parenthesis, equals 2 open parenthesis 1.12 close parenthesis to the power 2 t; equals 2 open bracket open parenthesis 1.12 close parenthesis squared close bracket to the power t equals 2 open parenthesis 1.2544 close parenthesis to the power t")[$P ( t ) & = 2 ( 1.12 )^(2 t) \ & = 2 attach([ ( 1.12 )^(2) ], t: t) = 2 ( 1.2544 )^(t)$] Thus, the annual growth factor for the price of butter is #math.equation(block: false, alt: "1.2544")[$1.2544$], and the annual percent growth rate is #math.equation(block: false, alt: "25.44")[$25.44$]%. In year0, the average annual cost of attending a public college was \$c0, and costs were climbing by r% per year. + Write a formula for #math.equation(block: false, alt: "C open parenthesis t close parenthesis")[$C ( t )$], the cost of one year of college #math.equation(block: false, alt: "t")[$t$] years after year0. #linebreak() #math.equation(block: false, alt: "C open parenthesis t close parenthesis equals")[$C ( t ) =$]\_\_\_\_\_ + Complete the table and sketch a graph of #math.equation(block: false, alt: "C open parenthesis t close parenthesis")[$C ( t )$]. Enter values without commas, rounded to the nearest whole dollar.#figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "10")[$10$]]), [#math.equation(block: false, alt: "C open parenthesis t close parenthesis")[$C ( t )$]], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], )) #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "25")[$25$]]), [#math.equation(block: false, alt: "C open parenthesis t close parenthesis")[$C ( t )$]], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], )) + If the percent growth rate remained steady, how much did a year of college cost in year1? + If the percent growth rate continues to remain steady, how much will a year of college cost in year2? #solutionbox[ + #math.equation(block: false, alt: "C open parenthesis t close parenthesis equals 10 , 069 times 1.06 to the power t")[$C ( t ) = 10 , 069 ⋅ 1.06^(t)$] + #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "10")[$10$]]), [#math.equation(block: false, alt: "C open parenthesis t close parenthesis")[$C ( t )$]], [cr0], [cr1], [cr2], )) #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "25")[$25$]]), [#math.equation(block: false, alt: "C open parenthesis t close parenthesis")[$C ( t )$]], [cr3], [cr4], [cr5], )) #linebreak() A graph is below. + \$cr1 per year + \$cr2 ] Another graph: #figure(figph[exponential growth], alt: "exponential growth", caption: none) In 2010, the average annual cost of tuition and fees at an in-state public college was \$7,130, and costs were climbing by 2.8% per year. + Write a formula for #math.equation(block: false, alt: "C open parenthesis t close parenthesis")[$C ( t )$], in thousands of dollars, the cost of one year of college #math.equation(block: false, alt: "t")[$t$] years after 2010. + Complete the table and sketch a graph of #math.equation(block: false, alt: "C open parenthesis t close parenthesis")[$C ( t )$]. #figure(table( columns: 7, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$" " t " "$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "25")[$25$]]), [#math.equation(block: false, alt: "C open parenthesis t close parenthesis")[$" " C ( t ) " "$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + If the percent growth rate remained steady, how much did a year of college cost in 2020? + If the percent growth rate continues to remain steady, how much will a year of college cost in 2030? #solutionbox[ + #math.equation(block: false, alt: "C open parenthesis t close parenthesis equals 7.130 times 1.028 to the power t")[$C ( t ) = 7.130 ⋅ 1.028^(t)$] + #figure(table( columns: 7, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$" " t " "$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "25")[$25$]]), [#math.equation(block: false, alt: "C open parenthesis t close parenthesis")[$" " C ( t ) " "$]], [7.130], [8.186], [9.398], [10.789], [12.387], [14.221], )) #figure(figph[exponential growth], alt: "exponential growth", caption: none) + \$9,398 + \$12,387 ] Describe carefully the meanings of #math.equation(block: false, alt: "b")[$b$] and #math.equation(block: false, alt: "r")[$r$] in the formulas for exponential growth. \_\_\_\_\_ Describe carefully the meanings of #math.equation(block: false, alt: "b")[$b$] and #math.equation(block: false, alt: "r")[$r$] in the formulas for exponential growth. === Exponential Decay In the preceding examples, exponential growth was modeled by increasing functions of the form #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals P sub 0 b to the power t")[$P ( t ) = P_(0) b^(t)$] where #math.equation(block: false, alt: "b greater than 1")[$b > 1$]. The function #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals P sub 0 b to the power t")[$P ( t ) = P_(0) b^(t)$] is a #emph[decreasing] function if #math.equation(block: false, alt: "0 less than b less than 1")[$0 < b < 1$]. In this case, we say that the function describes #strong[exponential decay], and the constant #math.equation(block: false, alt: "b")[$b$] is called the #strong[decay factor]. In Exponential Decay, we consider two examples of exponential decay. #notebox("Investigation", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Exponential Decay] + A small coal-mining town has been losing population since 1940, when 5000 people lived there. At each census thereafter (taken at 10-year intervals), the population declined to approximately 0.90 of its earlier figure.+ Fill in the table showing the population #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$] of the town #math.equation(block: false, alt: "t")[$t$] years after 1940.#figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$]], [], []), [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "5000")[$5000$]], [], [#math.equation(block: false, alt: "P open parenthesis 0 close parenthesis equals 5000")[$P ( 0 ) = 5000$]], [#math.equation(block: false, alt: "10")[$10$]], [], [], [#math.equation(block: false, alt: "P open parenthesis 10 close parenthesis equals 5000 times 0.90 equals")[$P ( 10 ) = 5000 ⋅ 0.90 =$]], [#math.equation(block: false, alt: "20")[$20$]], [], [], [#math.equation(block: false, alt: "P open parenthesis 20 close parenthesis equals open bracket 5000 times 0.90 close bracket times 0.90 equals")[$P ( 20 ) = [ 5000 ⋅ 0.90 ] ⋅ 0.90 =$]], [#math.equation(block: false, alt: "30")[$30$]], [], [], [#math.equation(block: false, alt: "P open parenthesis 3 close parenthesis equals")[$P ( 3 ) =$]], [#math.equation(block: false, alt: "40")[$40$]], [], [], [#math.equation(block: false, alt: "P open parenthesis 4 close parenthesis equals")[$P ( 4 ) =$]], [#math.equation(block: false, alt: "50")[$50$]], [], [], [#math.equation(block: false, alt: "P open parenthesis 5 close parenthesis equals")[$P ( 5 ) =$]], )) + Plot the data points and connect them with a smooth curve.#figure(figph[50x5000 grid], alt: "50x5000 grid", caption: none) + Write a function that gives the population of the town at any time #math.equation(block: false, alt: "t")[$t$] in years after 1940. #linebreak() Hint: Express the values you calculated in part (1) using powers of #math.equation(block: false, alt: "0.90")[$0.90$]. Do you see a connection between the value of #math.equation(block: false, alt: "t")[$t$] and the exponent on #math.equation(block: false, alt: "0.90")[$0.90$]? + Graph your function from part (3) using a calculator. (Use the table to choose an appropriate domain and range.) The graph should resemble your hand-drawn graph from part (2). + Evaluate your function to find the population of the town in 1995. What was the population in 2000? + A plastic window coating #math.equation(block: false, alt: "1")[$1$] millimeter thick decreases the light coming through a window by #math.equation(block: false, alt: "25")[$25$]%. This means that #math.equation(block: false, alt: "75")[$75$]% of the original amount of light comes through #math.equation(block: false, alt: "1")[$1$] millimeter of the coating. Each additional millimeter of coating reduces the light by another #math.equation(block: false, alt: "25")[$25$]%.+ Fill in the table showing the percent of the light, #math.equation(block: false, alt: "P open parenthesis x close parenthesis")[$P ( x )$], that shines through #math.equation(block: false, alt: "x")[$x$] millimeters of the window coating.#figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "P open parenthesis x close parenthesis")[$P ( x )$]], [], []), [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "100")[$100$]], [], [#math.equation(block: false, alt: "P open parenthesis 0 close parenthesis equals 100")[$P ( 0 ) = 100$]], [#math.equation(block: false, alt: "1")[$1$]], [], [], [#math.equation(block: false, alt: "P open parenthesis 1 close parenthesis equals 100 times 0.75 equals")[$P ( 1 ) = 100 ⋅ 0.75 =$]], [#math.equation(block: false, alt: "2")[$2$]], [], [], [#math.equation(block: false, alt: "P open parenthesis 2 close parenthesis equals open bracket 100 times 0.75 close bracket times 0.75 equals")[$P ( 2 ) = [ 100 ⋅ 0.75 ] ⋅ 0.75 =$]], [#math.equation(block: false, alt: "3")[$3$]], [], [], [#math.equation(block: false, alt: "P open parenthesis 3 close parenthesis equals")[$P ( 3 ) =$]], [#math.equation(block: false, alt: "4")[$4$]], [], [], [#math.equation(block: false, alt: "P open parenthesis 4 close parenthesis equals")[$P ( 4 ) =$]], [#math.equation(block: false, alt: "5")[$5$]], [], [], [#math.equation(block: false, alt: "P open parenthesis 5 close parenthesis equals")[$P ( 5 ) =$]], )) + Plot the data points and connect them with a smooth curve. #figure(figph[5x100 grid], alt: "5x100 grid", caption: none) + Write a function that gives the percent of the light that shines through #math.equation(block: false, alt: "x")[$x$] millimeters of the coating. #linebreak() Hint: Express the values you calculated in part (1) using powers of #math.equation(block: false, alt: "0.75")[$0.75$]. Do you see a connection between the value of #math.equation(block: false, alt: "x")[$x$] and the exponent on #math.equation(block: false, alt: "0.75")[$0.75$]? + Graph your function from part (3) using a calculator. (Use your table of values to choose an appropriate domain and range.) The graph should resemble your hand-drawn graph from part (2). + Evaluate your function to find the percent of the light that comes through 6 millimeters of plastic coating. What percent comes through #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$] millimeter? ] === Decay Factors Before Example, we noted that a percent increase of #math.equation(block: false, alt: "r")[$r$] (in decimal form) corresponds to a growth factor of #math.equation(block: false, alt: "b equals 1 plus r")[$b = 1 + r$]. A percent #emph[decrease] of #math.equation(block: false, alt: "r")[$r$] corresponds to a #emph[decay] factor of #math.equation(block: false, alt: "b equals 1 minus r")[$b = 1 − r$]. In Part B of Exponential Decay, each millimeter of plastic reduced the amount of light by #math.equation(block: false, alt: "25")[$25$]%, so #math.equation(block: false, alt: "r equals 0.25")[$r = 0.25$], and the decay factor for the function #math.equation(block: false, alt: "P open parenthesis x close parenthesis")[$P ( x )$] is #math.equation(block: true, alt: "b, equals 1 minus r; equals 1 minus 0.25 equals 0.75")[$b & = 1 − r \ & = 1 − 0.25 = 0.75$] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Note the difference in the two expressions for #math.equation(block: false, alt: "b")[$b$]: - A percent increase of #math.equation(block: false, alt: "r")[$r$] produces a growth factor of #math.equation(block: false, alt: "b equals 1 plus r")[$b = 1 + r$]. - A percent decrease of #math.equation(block: false, alt: "r")[$r$] produces a decay factor of #math.equation(block: false, alt: "b equals 1 minus r")[$b = 1 − r$]. ] A population decreases by 12% per year. What is its annual decay factor? \_\_\_\_\_ #solutionbox[ 0.88 ] A population decreases by 12% per year. What is its annual decay factor? + #math.equation(block: false, alt: "minus 0.12")[$− 0.12$] + #math.equation(block: false, alt: "minus 1.12")[$− 1.12$] + #math.equation(block: false, alt: "0.88")[$0.88$] + #math.equation(block: false, alt: "1.88")[$1.88$] #examplebox("Example 4")[][ David Reed writes in Context magazine: "Computing prices have been falling exponentially—50% every 18 months—for the past 30 years and will probably stay on that curve for another couple of decades." An accounting firm invests \$50,000 in new computer equipment. + Write a formula for the value of the equipment #math.equation(block: false, alt: "t")[$t$] years from now. + By what percent does the equipment depreciate each year? + What will the equipment be worth in #math.equation(block: false, alt: "5")[$5$] years? #solutionbox[ + The initial value of the equipment is #math.equation(block: false, alt: "V sub 0 equals 50 , 000")[$V_(0) = 50 , 000$]. Every #math.equation(block: false, alt: "18")[$18$] months, the value of the equipment is multiplied by #math.equation(block: true, alt: "b equals 1 minus r equals 1 minus 0.50 equals 0.50")[$b = 1 − r = 1 − 0.50 = 0.50$] However, because #math.equation(block: false, alt: "18")[$18$] months is #math.equation(block: false, alt: "1.5")[$1.5$] years, we must divide #math.equation(block: false, alt: "t")[$t$] by #math.equation(block: false, alt: "1.5")[$1.5$] in our formula, giving us #math.equation(block: true, alt: "V open parenthesis t close parenthesis equals 50 , 000 open parenthesis 0.50 close parenthesis to the power t / 1.5")[$V ( t ) = 50 , 000 ( 0.50 )^(t / 1.5)$] + After #math.equation(block: false, alt: "1")[$1$] year, we have #math.equation(block: true, alt: "V open parenthesis 1 close parenthesis equals 50 , 000 open parenthesis 0.50 close parenthesis to the power 1 / 1.5 equals 50 , 000 open parenthesis 0.63 close parenthesis")[$V ( 1 ) = 50 , 000 ( 0.50 )^(1 / 1.5) = 50 , 000 ( 0.63 )$] The equipment is worth #math.equation(block: false, alt: "63")[$63$]% of its original value, so it has depreciated by #math.equation(block: false, alt: "1 minus 0.63")[$1 − 0.63$], or #math.equation(block: false, alt: "37")[$37$]%. + After #math.equation(block: false, alt: "5")[$5$] years, #math.equation(block: true, alt: "V open parenthesis 5 close parenthesis equals 50 , 000 open parenthesis 0.50 close parenthesis to the power 5 / 1.5 equals 4960.628")[$V ( 5 ) = 50 , 000 ( 0.50 )^(5 / 1.5) = 4960.628$] To the nearest dollar, the equipment is worth \$4961. ] ] #figure(figph[grid], alt: "grid", caption: none) The number of butterflies visiting a nature station is declining by 18% per year. In year0, b0 butterflies visited the nature station. + What is the decay factor in the annual butterfly count? #linebreak() Answer: \_\_\_\_\_ + Write a formula for #math.equation(block: false, alt: "B open parenthesis t close parenthesis")[$B ( t )$], the number of butterflies #math.equation(block: false, alt: "t")[$t$] years after year0. #linebreak() #math.equation(block: false, alt: "B open parenthesis t close parenthesis equals")[$B ( t ) =$]\_\_\_\_\_ + Complete the table and sketch a graph of #math.equation(block: false, alt: "B open parenthesis t close parenthesis")[$B ( t )$]. A suggested grid is above.#figure(table( columns: 7, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "10")[$10$]]), [#math.equation(block: false, alt: "B open parenthesis t close parenthesis")[$B ( t )$]], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], )) #solutionbox[ + #math.equation(block: false, alt: "0.82")[$0.82$] + #math.equation(block: false, alt: "B open parenthesis t close parenthesis equals 3600 times 0.82 to the power t")[$B ( t ) = 3600 ⋅ 0.82^(t)$] + #figure(table( columns: 7, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "10")[$10$]]), [#math.equation(block: false, alt: "B open parenthesis t close parenthesis")[$B ( t )$]], [br0], [br1], [br2], [br3], [br4], [br5], )) #linebreak() A graph is below. ] #figure(figph[grid], alt: "grid", caption: none) The number of butterflies visiting a nature station is declining by 18% per year. In 1998, 3600 butterflies visited the nature station. + What is the decay factor in the annual butterfly count? + Write a formula for #math.equation(block: false, alt: "B open parenthesis t close parenthesis")[$B ( t )$], the number of butterflies #math.equation(block: false, alt: "t")[$t$] years after 1998. + Complete the table and sketch a graph of #math.equation(block: false, alt: "B open parenthesis t close parenthesis")[$B ( t )$].#figure(table( columns: 7, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "10")[$10$]]), [#math.equation(block: false, alt: "B open parenthesis t close parenthesis")[$B ( t )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) #solutionbox[ + #math.equation(block: false, alt: "0.82")[$0.82$] + #math.equation(block: false, alt: "B open parenthesis t close parenthesis equals 3600 times 0.82 to the power t")[$B ( t ) = 3600 ⋅ 0.82^(t)$] + #figure(table( columns: 7, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "10")[$10$]]), [#math.equation(block: false, alt: "B open parenthesis t close parenthesis")[$B ( t )$]], [3600], [2421], [1628], [1094], [736], [495], )) #figure(figph[decreasing exponential graph], alt: "decreasing exponential graph", caption: none) ] We summarize our observations about exponential growth and decay functions as follows. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Exponential Growth and Decay] The function #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals P sub 0 b to the power t")[$P ( t ) = P_(0) b^(t)$] models exponential growth and decay. - #math.equation(block: false, alt: "P sub 0 equals P open parenthesis 0 close parenthesis")[$P_(0) = P ( 0 )$] is the #strong[initial value] of #math.equation(block: false, alt: "P")[$P$]; - #math.equation(block: false, alt: "b")[$b$] is the #strong[growth] or #strong[decay factor]. + If #math.equation(block: false, alt: "b greater than 1")[$b > 1$], then #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$] is increasing, and #math.equation(block: false, alt: "b equals 1 plus r")[$b = 1 + r$], where #math.equation(block: false, alt: "r")[$r$] represents percent increase. + If #math.equation(block: false, alt: "0 less than b less than 1")[$0 < b < 1$], then #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$] is decreasing, and #math.equation(block: false, alt: "b equals 1 minus r")[$b = 1 − r$], where #math.equation(block: false, alt: "r")[$r$] represents percent decrease. ] The function #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals P sub 0 b to the power t")[$P ( t ) = P_(0) b^(t)$] is decreasing if \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "0 less than b less than 1")[$0 < b < 1$] ] The function #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals P sub 0 b to the power t")[$P ( t ) = P_(0) b^(t)$] is decreasing if + #math.equation(block: false, alt: "P sub 0")[$P_(0)$] is negative. + #math.equation(block: false, alt: "b")[$b$] is negative. + #math.equation(block: false, alt: "0 less than b less than 1")[$0 < b < 1$] + #math.equation(block: false, alt: "0 less than P sub 0 less than 1")[$0 < P_(0) < 1$] === Comparing Linear Growth and Exponential Growth It may be helpful to compare linear growth and exponential growth. Consider the two functions #math.equation(block: true, alt: "L open parenthesis t close parenthesis equals 5 plus 2 t and E open parenthesis t close parenthesis equals 5 times 2 to the power t open parenthesis t greater than or equal to 0 close parenthesis")[$L ( t ) = 5 + 2 t " " " and " " " E ( t ) = 5 ⋅ 2^(t) " " " " " " ( t ≥ 0 )$] whose graphs are shown below. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "L open parenthesis t close parenthesis")[$L ( t )$]]), [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "11")[$11$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "13")[$13$]], )) Slope #math.equation(block: false, alt: "m equals 2")[$m = 2$] #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "E open parenthesis t close parenthesis")[$E ( t )$]]), [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "80")[$80$]], )) Growth factor #math.equation(block: false, alt: "b equals 2")[$b = 2$] #figure(figph[graph of linear and exponential], alt: "graph of linear and exponential", caption: none) #math.equation(block: false, alt: "L")[$L$] is a linear function with initial value #math.equation(block: false, alt: "5")[$5$] and slope #math.equation(block: false, alt: "2")[$2$]; #math.equation(block: false, alt: "E")[$E$] is an exponential function with initial value #math.equation(block: false, alt: "5")[$5$] and growth factor #math.equation(block: false, alt: "2")[$2$]. In a way, the growth factor of an exponential function is analogous to the slope of a linear function: Each measures how quickly the function is increasing (or decreasing). However, for each unit increase in #math.equation(block: false, alt: "t")[$t$], #math.equation(block: false, alt: "2")[$2$] units are #emph[added] to the value of #math.equation(block: false, alt: "L open parenthesis t close parenthesis")[$L ( t )$], whereas the value of #math.equation(block: false, alt: "E open parenthesis t close parenthesis")[$E ( t )$] is #emph[multiplied] by #math.equation(block: false, alt: "2")[$2$]. An exponential function with growth factor #math.equation(block: false, alt: "2")[$2$] eventually grows much more rapidly than a linear function with slope #math.equation(block: false, alt: "2")[$2$], as you can see by comparing the graphs in the figure or the function values in the tables. The graph of #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals P sub 0 b to the power t")[$P ( t ) = P_(0) b^(t)$], where #math.equation(block: false, alt: "P sub 0 greater than 0")[$P_(0) > 0$], #math.equation(block: false, alt: "b greater than 0")[$b > 0$], and #math.equation(block: false, alt: "b not equal to 1")[$b ≠ 1$], is always \_\_\_\_\_ #solutionbox[ concave up ] The graph of #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals P sub 0 b to the power t")[$P ( t ) = P_(0) b^(t)$], where #math.equation(block: false, alt: "P sub 0 greater than 0")[$P_(0) > 0$], #math.equation(block: false, alt: "b greater than 0")[$b > 0$], and #math.equation(block: false, alt: "b not equal to 1")[$b ≠ 1$] is always + increasing. + concave up. + decreasing. + concave down. #examplebox("Example 5")[][ A solar energy company sold \$#math.equation(block: false, alt: "80 , 000")[$80 , 000$] worth of solar collectors last year, its first year of operation. This year its sales rose to \$#math.equation(block: false, alt: "88 , 000")[$88 , 000$], an increase of #math.equation(block: false, alt: "10")[$10$]%. The marketing department must estimate its projected sales for the next #math.equation(block: false, alt: "3")[$3$] years. + If the marketing department predicts that sales will grow linearly, what should it expect the sales total to be next year? Graph the projected sales figures over the next #math.equation(block: false, alt: "3")[$3$] years, assuming that sales will grow linearly. + If the marketing department predicts that sales will grow exponentially, what should it expect the sales total to be next year? Graph the projected sales figures over the next #math.equation(block: false, alt: "3")[$3$] years, assuming that sales will grow exponentially. #solutionbox[ + Let #math.equation(block: false, alt: "L open parenthesis t close parenthesis")[$L ( t )$] represent the company's total sales #math.equation(block: false, alt: "t")[$t$] years after starting business, where #math.equation(block: false, alt: "t equals 0")[$t = 0$] is the first year of operation. If sales grow linearly, then #math.equation(block: false, alt: "L open parenthesis t close parenthesis")[$L ( t )$] has the form #math.equation(block: false, alt: "L open parenthesis t close parenthesis equals b plus m t .")[$L ( t ) = b + m t .$] Now #math.equation(block: false, alt: "L open parenthesis 0 close parenthesis equals 80 , 000")[$L ( 0 ) = 80 , 000$], so the intercept #math.equation(block: false, alt: "b")[$b$] is #math.equation(block: false, alt: "80 , 000")[$80 , 000$]. The slope #math.equation(block: false, alt: "m")[$m$] of the graph is #math.equation(block: true, alt: "the fraction Δ S over Δ t equals the fraction 8000 dollars over 1 year equals 8000 dollars/year")[$frac(upright(Δ) S, upright(Δ) t) = frac(8000 " dollars", 1 " year") = 8000 " dollars/year"$] where #math.equation(block: false, alt: "Δ S equals 8000")[$upright(Δ) S = 8000$] is the increase in sales during the first year. Thus, #math.equation(block: false, alt: "L open parenthesis t close parenthesis equals 80 , 000 plus 8000 t ,")[$L ( t ) = 80 , 000 + 8000 t ,$] and sales grow by adding \$#math.equation(block: false, alt: "8000")[$8000$] each year. The expected sales total for the next year is #math.equation(block: true, alt: "L open parenthesis 2 close parenthesis equals 80 , 000 plus 8000 open parenthesis 2 close parenthesis equals 96 , 000")[$L ( 2 ) = 80 , 000 + 8000 ( 2 ) = 96 , 000$] + Let #math.equation(block: false, alt: "E open parenthesis t close parenthesis")[$E ( t )$] represent the company's sales assuming that sales will grow exponentially. Then #math.equation(block: false, alt: "E open parenthesis t close parenthesis")[$E ( t )$] has the form #math.equation(block: false, alt: "E open parenthesis t close parenthesis equals E sub 0 b to the power t")[$E ( t ) = E_(0) b^(t)$] . The percent increase in sales over the first year was #math.equation(block: false, alt: "r equals 0.10")[$r = 0.10$], so the growth rate is #math.equation(block: true, alt: "b equals 1 plus r equals 1.10")[$b = 1 + r = 1.10$] The initial value, #math.equation(block: false, alt: "E sub 0")[$E_(0)$], is #math.equation(block: false, alt: "80 , 000")[$80 , 000$]. Thus, #math.equation(block: false, alt: "E open parenthesis t close parenthesis equals 80 , 000 open parenthesis 1.10 close parenthesis to the power t")[$E ( t ) = 80 , 000 ( 1.10 )^(t)$], and sales grow by being multiplied each year by #math.equation(block: false, alt: "1.10")[$1.10$]. The expected sales total for the next year is #math.equation(block: true, alt: "E open parenthesis 2 close parenthesis equals 80 , 000 open parenthesis 1.10 close parenthesis squared equals 96 , 800")[$E ( 2 ) = 80 , 000 ( 1.10 )^(2) = 96 , 800$] We evaluate each function at several points to obtain the graphs shown in the figure. #figure(table( columns: 3, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "L open parenthesis t close parenthesis")[$L ( t )$]], [#math.equation(block: false, alt: "E open parenthesis t close parenthesis")[$E ( t )$]]), [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "80 , 000")[$80 , 000$]], [#math.equation(block: false, alt: "80 , 000")[$80 , 000$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "88 , 000")[$88 , 000$]], [#math.equation(block: false, alt: "88 , 000")[$88 , 000$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "96 , 000")[$96 , 000$]], [#math.equation(block: false, alt: "96 , 800")[$96 , 800$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "104 , 000")[$104 , 000$]], [#math.equation(block: false, alt: "106 , 480")[$106 , 480$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "112 , 000")[$112 , 000$]], [#math.equation(block: false, alt: "117 , 128")[$117 , 128$]], )) #figure(figph[linear and exponential growth], alt: "linear and exponential growth", caption: none) ] ] A new car begins to depreciate in value as soon as you drive it off the lot. Some models depreciate linearly, and others depreciate exponentially. Suppose you buy a new car for \$20,000, and #math.equation(block: false, alt: "1")[$1$] year later its value has decreased to \$17,000. + If the value decreased linearly, what was its annual rate of decrease? #linebreak() \$\_\_\_\_\_ per year + If the value decreased exponentially, what was its annual decay factor? \_\_\_\_\_ What was its annual percent depreciation? \_\_\_\_\_% + Calculate the value of your car when it is #math.equation(block: false, alt: "5")[$5$] years old under each assumption, linear or exponential depreciation. #linebreak() Linear: \$\_\_\_\_\_ #linebreak() Exponential: \$\_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "$ 3000")[$upright(\$) 3000$] per year + #math.equation(block: false, alt: "0.85")[$0.85$]; #math.equation(block: false, alt: "15 %")[$15 upright(%)$] + Linear: #math.equation(block: false, alt: "$ 5000")[$upright(\$) 5000$]; Exponential: #math.equation(block: false, alt: "$ 8874")[$upright(\$) 8874$] ] A new car begins to depreciate in value as soon as you drive it off the lot. Some models depreciate linearly, and others depreciate exponentially. Suppose you buy a new car for \$20,000, and #math.equation(block: false, alt: "1")[$1$] year later its value has decreased to \$17,000. + If the value decreased linearly, what was its annual rate of decrease? + If the value decreased exponentially, what was its annual decay factor? What was its annual percent depreciation? + Calculate the value of your car when it is #math.equation(block: false, alt: "5")[$5$] years old under each assumption, linear or exponential depreciation. #solutionbox[ + #math.equation(block: false, alt: "$ 3000")[$upright(\$) 3000$] per year + #math.equation(block: false, alt: "0.85")[$0.85$]; #math.equation(block: false, alt: "15 %")[$15 upright(%)$] + Linear: #math.equation(block: false, alt: "$ 5000")[$upright(\$) 5000$]; Exponential: #math.equation(block: false, alt: "$ 8874")[$upright(\$) 8874$] ] Explain the difference between constant slope and constant growth factor. \_\_\_\_\_ Explain the difference between constant slope and constant growth factor. === Section Summary ==== Vocabulary Look up the definitions of new terms in the Glossary. - Exponential growth - Initial value - Exponential decay - Percent increase - Compound interest - Growth factor - Amount ==== CONCEPTS + If a quantity is multiplied by a constant factor, #math.equation(block: false, alt: "b")[$b$], in each time period, we say that it undergoes #strong[exponential growth] or #strong[decay]. The constant #math.equation(block: false, alt: "b")[$b$] is called the #strong[growth factor] if #math.equation(block: false, alt: "b greater than 1")[$b > 1$] and the #strong[decay factor] if #math.equation(block: false, alt: "0 less than b less than 1")[$0 < b < 1$]. + Quantities that increase or decrease by a constant percent in each time period grow or decay exponentially. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Exponential Growth and Decay] The function #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals P sub 0 b to the power t")[$P ( t ) = P_(0) b^(t)$]models exponential growth and decay. #math.equation(block: false, alt: "P sub 0 equals P open parenthesis 0 close parenthesis")[$P_(0) = P ( 0 )$] is the #strong[initial value] of #math.equation(block: false, alt: "P")[$P$]; #math.equation(block: false, alt: "b")[$b$] is the #strong[growth] or #strong[decay factor]. + If #math.equation(block: false, alt: "b greater than 1")[$b > 1$], then #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$] is increasing, and #math.equation(block: false, alt: "b equals 1 plus r")[$b = 1 + r$], where #math.equation(block: false, alt: "r")[$r$] represents percent increase. + If #math.equation(block: false, alt: "0 less than b less than 1")[$0 < b < 1$], then #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$] is decreasing, and #math.equation(block: false, alt: "b equals 1 minus r")[$b = 1 − r$], where #math.equation(block: false, alt: "r")[$r$] represents percent decrease. ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Compound Interest] The amount #math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$] accumulated (principal plus interest) in an account bearing interest compounded annually is #math.equation(block: true, alt: "A open parenthesis t close parenthesis equals P open parenthesis 1 plus r close parenthesis to the power t")[$A ( t ) = P ( 1 + r )^(t)$]where #figure(table( columns: 3, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "P")[$P$]], [], [is the principal invested,]), [#math.equation(block: false, alt: "r")[$r$]], [], [is the interest rate,], [#math.equation(block: false, alt: "t")[$t$]], [], [is the time period, in years], )) ] + In linear growth, a constant amount is #emph[added] to the output for each unit increase in the input. In exponential growth, the output is #emph[multiplied] by a constant factor for each unit increase in the input. ==== STUDY QUESTIONS + Is it possible for two populations with the same initial value to grow at different percent rates? + If you know the percent growth rate, how can you find the growth factor? If you know the percent decay rate, how can you find the decay factor? + What is the growth factor for a population that grows #math.equation(block: false, alt: "4 %")[$4 upright(%)$] annually? + What is the decay factor for a population that declines by #math.equation(block: false, alt: "4 %")[$4 upright(%)$] annually? + What is the growth factor for a population that grows by #math.equation(block: false, alt: "100 %")[$100 upright(%)$] annually? + Explain the difference between the slope in linear growth and the growth factor in exponential growth. ==== SKILLS Practice each skill in the Homework problems listed. + Calculate percent increase or decrease: \#1–10–6 + Write a formula for exponential growth or decay: \#11–22 + Evaluate an exponential growth or decay function: \#11–22 + Simplify exponential expressions: \#23–32 + Solve power equations: \#33–40 + Find the growth factor or initial value: \#41–58 + Solve for percent increase or decrease: \#63–66 === Homework 4.1 + A parking permit at Huron College cost \$#math.equation(block: false, alt: "25")[$25$] last year, but this year the price increased by #math.equation(block: false, alt: "12 %")[$12 upright(%)$]. What is the price this year? + If the price of a parking permit increases by #math.equation(block: false, alt: "12 %")[$12 upright(%)$] again next year, what will the price be then? + \$#math.equation(block: false, alt: "28")[$28$] + \$#math.equation(block: false, alt: "31.36")[$31.36$] + The computer you want cost \$#math.equation(block: false, alt: "1200")[$1200$] when it first came on the market, but after #math.equation(block: false, alt: "3")[$3$] months the price was reduced by #math.equation(block: false, alt: "15 %")[$15 upright(%)$]. What was the price then? + If the price falls by another #math.equation(block: false, alt: "15 %")[$15 upright(%)$] next month, what will the price be then? The value of your stock portfolio fell #math.equation(block: false, alt: "10 %")[$10 upright(%)$] last year, but this year it increased by #math.equation(block: false, alt: "10 %")[$10 upright(%)$]. How does the current value of your portfolio compare to what it was two years ago? It is #math.equation(block: false, alt: "99 %")[$99 upright(%)$] of what it was #math.equation(block: false, alt: "2")[$2$] years ago. You got a #math.equation(block: false, alt: "5 %")[$5 upright(%)$] raise in January, but then in March everyone took a pay cut of #math.equation(block: false, alt: "5 %")[$5 upright(%)$]. How does your new salary compare to what it was last December? The population of Summerville is currently #math.equation(block: false, alt: "12")[$12$] hundred people. + Write a formula for the population if it grows at a constant rate of #math.equation(block: false, alt: "1.5")[$1.5$] hundred people per year. What is the population after #math.equation(block: false, alt: "3")[$3$] years? + Write a formula for the population if it has a constant growth factor of #math.equation(block: false, alt: "1.5")[$1.5$] per year. What is the population after #math.equation(block: false, alt: "3")[$3$] years? + #math.equation(block: false, alt: "P equals 1200 plus 150 t")[$P = 1200 + 150 t$]; #math.equation(block: false, alt: "1650")[$1650$] + #math.equation(block: false, alt: "P equals 1200 times 1.5 to the power t")[$P = 1200 ⋅ 1.5^(t)$]; #math.equation(block: false, alt: "4050")[$4050$] Delbert's sports car was worth \$#math.equation(block: false, alt: "45 , 000")[$45 , 000$] when he bought it. + Write a formula for the value of the car if it depreciates at a constant rate of \$#math.equation(block: false, alt: "7000")[$7000$] per year. What is the value of the car after #math.equation(block: false, alt: "4")[$4$] years? + Write a formula for the value of the car if it has a constant depreciation factor of #math.equation(block: false, alt: "0.70")[$0.70$] per year. What is the value of the car after #math.equation(block: false, alt: "4")[$4$] years? Francine's truck was worth \$#math.equation(block: false, alt: "18 , 000")[$18 , 000$] when she bought it. + Write a formula for the value of the truck if it depreciates by \$#math.equation(block: false, alt: "2000")[$2000$] per year. What is the value of the truck after #math.equation(block: false, alt: "5")[$5$] years? + Write a formula for the value of the truck if it depreciates by #math.equation(block: false, alt: "20 %")[$20 upright(%)$] per year. What is the value of the truck after #math.equation(block: false, alt: "5")[$5$] years? + #math.equation(block: false, alt: "V equals 18 , 000 minus 2000 t")[$V = 18 , 000 − 2000 t$]; \$#math.equation(block: false, alt: "8000")[$8000$] + #math.equation(block: false, alt: "V equals 18 , 000 times 0.8 to the power t")[$V = 18 , 000 ⋅ 0.8^(t)$]; \$#math.equation(block: false, alt: "5898.24")[$5898.24$] The population of Lakeview is currently #math.equation(block: false, alt: "150 , 000")[$150 , 000$] people. + Write a formula for the population if it grows by #math.equation(block: false, alt: "6000")[$6000$] people per year. What is the population after #math.equation(block: false, alt: "2")[$2$] years? + Write a formula for the population if grows by #math.equation(block: false, alt: "6 %")[$6 upright(%)$] per year. What is the population after #math.equation(block: false, alt: "2")[$2$] years? The table shows the growth factor for a number of different populations. For each population, find the percent growth rate. #figure(table( columns: 6, align: left, inset: 6pt, table.header([Population], [#math.equation(block: false, alt: "A")[$A$]], [#math.equation(block: false, alt: "B")[$B$]], [#math.equation(block: false, alt: "C")[$C$]], [#math.equation(block: false, alt: "D")[$D$]], [#math.equation(block: false, alt: "E")[$E$]]), [Growth factor], [#math.equation(block: false, alt: "1.2")[$1.2$]], [#math.equation(block: false, alt: "1.02")[$1.02$]], [#math.equation(block: false, alt: "1.075")[$1.075$]], [#math.equation(block: false, alt: "2.0")[$2.0$]], [#math.equation(block: false, alt: "2.15")[$2.15$]], [Percent growth rate], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$], )) A: #math.equation(block: false, alt: "20 %")[$20 upright(%)$]; B: #math.equation(block: false, alt: "2 %")[$2 upright(%)$]; C: #math.equation(block: false, alt: "7.5 %")[$7.5 upright(%)$]; D: #math.equation(block: false, alt: "100 %")[$100 upright(%)$]; E: #math.equation(block: false, alt: "115 %")[$115 upright(%)$] The table shows the decay factor for a number of different populations. For each population, find the percent decay rate. #figure(table( columns: 6, align: left, inset: 6pt, table.header([Population], [#math.equation(block: false, alt: "A")[$A$]], [#math.equation(block: false, alt: "B")[$B$]], [#math.equation(block: false, alt: "C")[$C$]], [#math.equation(block: false, alt: "D")[$D$]], [#math.equation(block: false, alt: "E")[$E$]]), [Decay factor], [#math.equation(block: false, alt: "0.6")[$0.6$]], [#math.equation(block: false, alt: "0.06")[$0.06$]], [#math.equation(block: false, alt: "0.96")[$0.96$]], [#math.equation(block: false, alt: "0.996")[$0.996$]], [#math.equation(block: false, alt: "0.096")[$0.096$]], [Percent decay rate], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$], [$#hide($00000$)$], )) For Problems 11–16, + Write a function that describes exponential growth. + Graph the function. + Evaluate the function at the given values. A typical beehive contains #math.equation(block: false, alt: "20 , 000")[$20 , 000$] insects. The population can increase in size by a factor of #math.equation(block: false, alt: "2.5")[$2.5$] every #math.equation(block: false, alt: "6")[$6$] weeks. How many bees could there be after #math.equation(block: false, alt: "4")[$4$] weeks? After #math.equation(block: false, alt: "20")[$20$] weeks? + #math.equation(block: false, alt: "P equals 20 , 000 times 2.5 to the power t / 6")[$P = 20 , 000 ⋅ 2.5^(t / 6)$] + #figure(figph[exponential growth], alt: "exponential growth", caption: none) + #math.equation(block: false, alt: "36 , 840")[$36 , 840$] bees; #math.equation(block: false, alt: "424 , 128")[$424 , 128$] bees A rancher who started with #math.equation(block: false, alt: "800")[$800$] head of cattle finds that his herd increases by a factor of #math.equation(block: false, alt: "1.8")[$1.8$] every #math.equation(block: false, alt: "3")[$3$] years. How many head of cattle will he have after #math.equation(block: false, alt: "1")[$1$] year? After #math.equation(block: false, alt: "10")[$10$] years? A sum of \$#math.equation(block: false, alt: "4000")[$4000$] is invested in an account that pays #math.equation(block: false, alt: "8 %")[$8 upright(%)$] interest compounded annually. How much is in the account after #math.equation(block: false, alt: "2")[$2$] years? After #math.equation(block: false, alt: "10")[$10$] years? + #math.equation(block: false, alt: "A equals 4000 times 1.08 to the power t")[$A = 4000 ⋅ 1.08^(t)$] + #figure(figph[exponential growth], alt: "exponential growth", caption: none) + \$#math.equation(block: false, alt: "4665.60")[$4665.60$]; \$#math.equation(block: false, alt: "8635.70")[$8635.70$] Otto invests \$#math.equation(block: false, alt: "600")[$600$] in an account that pays #math.equation(block: false, alt: "7.3 %")[$7.3 upright(%)$] interest compounded annually. How much is in Otto's account after #math.equation(block: false, alt: "3")[$3$] years? After #math.equation(block: false, alt: "6")[$6$] years? Paul bought a house for \$#math.equation(block: false, alt: "200 , 000")[$200 , 000$] in #math.equation(block: false, alt: "2003")[$2003$]. Since #math.equation(block: false, alt: "2003")[$2003$], housing prices have risen an average of #math.equation(block: false, alt: "5 %")[$5 upright(%)$] per year. How much was the house worth in #math.equation(block: false, alt: "2015")[$2015$]? How much will it be worth in 2030? + #math.equation(block: false, alt: "P equals 200 , 000 times 1.05 to the power t")[$P = 200 , 000 ⋅ 1.05^(t)$] + #figure(figph[exponential growth], alt: "exponential growth", caption: none) + \$#math.equation(block: false, alt: "359 , 171")[$359 , 171$]; \$#math.equation(block: false, alt: "746 , 691")[$746 , 691$] Sales of Windsurfers have increased #math.equation(block: false, alt: "12 %")[$12 upright(%)$] per year since #math.equation(block: false, alt: "2010")[$2010$]. If Sunsails sold #math.equation(block: false, alt: "1500")[$1500$] Windsurfers in #math.equation(block: false, alt: "2010")[$2010$], how many did it sell in #math.equation(block: false, alt: "2015")[$2015$]? How many should it expect to sell in #math.equation(block: false, alt: "2022")[$2022$]? For Problems 17–22, + Write a function that describes exponential decay. + Graph the function. + Evaluate the function at the given values. During a vigorous spraying program, the mosquito population was reduced to #math.equation(block: false, alt: "the fraction 3 over 4")[$display(frac(3, 4))$] of its previous size every #math.equation(block: false, alt: "2")[$2$] weeks. If the mosquito population was originally estimated at #math.equation(block: false, alt: "250 , 000")[$250 , 000$], how many mosquitoes remained after #math.equation(block: false, alt: "3")[$3$] weeks of spraying? After #math.equation(block: false, alt: "8")[$8$] weeks? + #math.equation(block: false, alt: "P equals 250 , 000 times 0.75 to the power t / 2")[$P = 250 , 000 ⋅ 0.75^(t / 2)$] + #figure(figph[exponential decay], alt: "exponential decay", caption: none) + #math.equation(block: false, alt: "162 , 380")[$162 , 380$]; #math.equation(block: false, alt: "79 , 102")[$79 , 102$] The number of perch in Hidden Lake has declined to half of its previous value every #math.equation(block: false, alt: "5")[$5$] years since 1985, when the perch population was estimated at #math.equation(block: false, alt: "8000")[$8000$]. How many perch were there in 1995? In 2013? Scuba divers find that the water in Emerald Lake filters out #math.equation(block: false, alt: "15 %")[$15 upright(%)$] of the sunlight for each #math.equation(block: false, alt: "4")[$4$] feet that they descend. How much sunlight penetrates to a depth of #math.equation(block: false, alt: "20")[$20$] feet? To a depth of #math.equation(block: false, alt: "45")[$45$] feet? + #math.equation(block: false, alt: "L equals 0.85 to the power d / 4")[$L = 0.85^(d / 4)$] + #figure(figph[exponential decay], alt: "exponential decay", caption: none) + #math.equation(block: false, alt: "44 %")[$44 upright(%)$]; #math.equation(block: false, alt: "16 %")[$16 upright(%)$] Arch's motorboat cost \$#math.equation(block: false, alt: "15 , 000")[$15 , 000$] in #math.equation(block: false, alt: "2005")[$2005$] and has depreciated by #math.equation(block: false, alt: "10 %")[$10 upright(%)$] every #math.equation(block: false, alt: "3")[$3$] years. How much was the boat worth in #math.equation(block: false, alt: "2014")[$2014$]? In #math.equation(block: false, alt: "2015")[$2015$]? Plutonium-238 is a radioactive element that decays over time into a less harmful element at a rate of #math.equation(block: false, alt: "0.8 %")[$0.8 upright(%)$] per year. A power plant has #math.equation(block: false, alt: "50")[$50$] pounds of plutonium-238 to dispose of. How much plutonium-238 will be left after #math.equation(block: false, alt: "10")[$10$] years? After #math.equation(block: false, alt: "100")[$100$] years? + #math.equation(block: false, alt: "P equals 50 times 0.992 to the power t")[$P = 50 ⋅ 0.992^(t)$] + #figure(figph[exponential decay], alt: "exponential decay", caption: none) + #math.equation(block: false, alt: "46.1")[$46.1$] lb; #math.equation(block: false, alt: "22.4")[$22.4$] lb Iodine-131 is a radioactive element that decays at a rate of #math.equation(block: false, alt: "8.3 %")[$8.3 upright(%)$] per day. How much of a #math.equation(block: false, alt: "12")[$12$]-gram sample will be left after #math.equation(block: false, alt: "1")[$1$] week? After #math.equation(block: false, alt: "15")[$15$] days? In Problems 23–26, use the laws of exponents to simplify. + #math.equation(block: false, alt: "3 to the power x 3 to the power 4")[$3^(x) #h(0.167em) 3^(4)$] + #math.equation(block: false, alt: "open parenthesis 3 to the power x close parenthesis to the power 4")[$( 3^(x) )^(4)$] + #math.equation(block: false, alt: "3 to the power x 4 to the power x")[$3^(x) #h(0.167em) 4^(x)$] + #math.equation(block: false, alt: "3 to the power x plus 4")[$3^(x + 4)$] + #math.equation(block: false, alt: "3 to the power 4 x")[$3^(4 x)$] + #math.equation(block: false, alt: "12 to the power x")[$12^(x)$] + #math.equation(block: false, alt: "8 to the power x 8 to the power x")[$8^(x) #h(0.167em) 8^(x)$] + #math.equation(block: false, alt: "8 to the power x plus 28 to the power x minus 1")[$8^(x + 2) #h(0.167em) 8^(x − 1)$] + #math.equation(block: false, alt: "the fraction 8 to the power 2 x over 8 to the power x")[$display(frac(8^(2 x), 8^(x)))$] + #math.equation(block: false, alt: "b to the power minus 4 t b to the power 2 t")[$b^(− 4 t) #h(0.167em) b^(2 t)$] + #math.equation(block: false, alt: "open parenthesis b to the power t close parenthesis to the power 1 / 2")[$( b^(t) )^(1 / 2)$] + #math.equation(block: false, alt: "b to the power t minus 1 b to the power 1 minus t")[$b^(t − 1) #h(0.167em) b^(1 − t)$] + #math.equation(block: false, alt: "b to the power minus 2 t")[$b^(− 2 t)$] + #math.equation(block: false, alt: "b to the power t / 2")[$b^(t / 2)$] + #math.equation(block: false, alt: "1")[$1$] + #math.equation(block: false, alt: "b to the power t / 2 b to the power t / 2")[$b^(t / 2) #h(0.167em) b^(t / 2)$] + #math.equation(block: false, alt: "the fraction b to the power 2 t over b")[$display(frac(b^(2 t), b))$] + #math.equation(block: false, alt: "b to the power 1 / t b to the power t")[$b^(1 / t) #h(0.167em) b^(t)$] Let #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals 12 open parenthesis 3 close parenthesis to the power t")[$P ( t ) = 12 ( 3 )^(t)$]. Show that #math.equation(block: false, alt: "P open parenthesis t plus 1 close parenthesis equals 3 P open parenthesis t close parenthesis")[$P ( t + 1 ) = 3 P ( t )$]. #math.equation(block: true, alt: "P open parenthesis t plus 1 close parenthesis equals 12 open parenthesis 3 close parenthesis to the power t plus 1 equals 12 open parenthesis 3 close parenthesis to the power t times 3 equals P open parenthesis t close parenthesis times 3")[$P ( t + 1 ) = 12 ( 3 )^(t + 1) = 12 ( 3 )^(t) ⋅ 3 = P ( t ) ⋅ 3$] Let #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals 8 open parenthesis 5 close parenthesis to the power t")[$N ( t ) = 8 ( 5 )^(t)$]. Show that #math.equation(block: false, alt: "the fraction N open parenthesis t plus k close parenthesis over N open parenthesis t close parenthesis equals 5 to the power k")[$display(frac(N ( t + k ), N ( t ))) = 5^(k)$] Let #math.equation(block: false, alt: "P open parenthesis x close parenthesis equals P sub 0 a to the power x")[$P ( x ) = P_(0) a^(x)$]. Show that #math.equation(block: false, alt: "P open parenthesis x plus k close parenthesis equals a to the power k P open parenthesis x close parenthesis")[$P ( x + k ) = a^(k) P ( x )$]. #math.equation(block: true, alt: "P open parenthesis x plus k close parenthesis equals P sub 0 a to the power x plus k equals P sub 0 a to the power x times a to the power k equals P open parenthesis x close parenthesis times a to the power k")[$P ( x + k ) = P_(0) a^(x + k) = P_(0) a^(x) ⋅ a^(k) = P ( x ) ⋅ a^(k)$] Let #math.equation(block: false, alt: "N open parenthesis x close parenthesis equals N sub 0 b to the power x")[$N ( x ) = N_(0) b^(x)$]. Show that #math.equation(block: false, alt: "the fraction N open parenthesis x plus 1 close parenthesis over N open parenthesis x close parenthesis equals b")[$display(frac(N ( x + 1 ), N ( x ))) = b$] + Explain why #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals 2 times 3 to the power t")[$P ( t ) = 2 ⋅ 3^(t)$] and #math.equation(block: false, alt: "Q open parenthesis t close parenthesis equals 6 to the power t")[$Q ( t ) = 6^(t)$] are not the same function. + Complete the table of values for #math.equation(block: false, alt: "P")[$P$] and #math.equation(block: false, alt: "Q")[$Q$], showing that their values are not the same.#figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]]), [#math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [#math.equation(block: false, alt: "Q open parenthesis t close parenthesis")[$Q ( t )$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + In the expression #math.equation(block: false, alt: "2 times 3 to the power t")[$2 ⋅ 3^(t)$], only the #math.equation(block: false, alt: "3")[$3$] is raised to a power #math.equation(block: false, alt: "t")[$t$], and the result is doubled, but if both the #math.equation(block: false, alt: "2")[$2$] and the #math.equation(block: false, alt: "3")[$3$] were raised to the power #math.equation(block: false, alt: "t")[$t$], the result would be #math.equation(block: false, alt: "6 to the power t")[$6^(t)$]. + #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]]), [#math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "18")[$18$]], [#math.equation(block: false, alt: "Q open parenthesis t close parenthesis")[$Q ( t )$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "36")[$36$]], )) + Explain why #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals 4 times open parenthesis the fraction 1 over 2 close parenthesis to the power t")[$P ( t ) = 4 ⋅ attach(( display(frac(1, 2)) ), t: t)$] and #math.equation(block: false, alt: "Q open parenthesis t close parenthesis equals 2 to the power t")[$Q ( t ) = 2^(t)$] are not the same function. + Complete the table of values for #math.equation(block: false, alt: "P")[$P$] and #math.equation(block: false, alt: "Q")[$Q$], showing that their values are not the same.#figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]]), [#math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [#math.equation(block: false, alt: "Q open parenthesis t close parenthesis")[$Q ( t )$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) Solve the equation. (See Roots and Radicals to review solving equations involving powers of the variable.) Round your answer to two places if necessary. #math.equation(block: true, alt: "768 equals 12 b cubed")[$768 = 12 b^(3)$] #math.equation(block: true, alt: "4")[$4$] #math.equation(block: true, alt: "75 equals 3 b to the power 4")[$75 = 3 b^(4)$] #math.equation(block: true, alt: "14 , 929.92 equals 5000 b to the power 6")[$14 , 929.92 = 5000 b^(6)$] #math.equation(block: true, alt: "1.2")[$1.2$] #math.equation(block: true, alt: "151 , 875 equals 20 , 000 b to the power 5")[$151 , 875 = 20 , 000 b^(5)$] #math.equation(block: true, alt: "1253 equals 260 open parenthesis 1 plus r close parenthesis to the power 12")[$1253 = 260 ( 1 + r )^(12)$] #math.equation(block: true, alt: "r approximately equals 0.14")[$r ≈ 0.14$] #math.equation(block: true, alt: "116 , 473 equals 48 , 600 open parenthesis 1 plus r close parenthesis to the power 15")[$116 , 473 = 48 , 600 ( 1 + r )^(15)$] #math.equation(block: true, alt: "56.27 equals 78 open parenthesis 1 minus r close parenthesis to the power 8")[$56.27 = 78 ( 1 − r )^(8)$] #math.equation(block: true, alt: "r approximately equals 0.04")[$r ≈ 0.04$] #math.equation(block: true, alt: "10.56 equals 12.4 open parenthesis 1 minus r close parenthesis to the power 20")[$10.56 = 12.4 ( 1 − r )^(20)$] + Riverside County is the fastest growing county in California. In #math.equation(block: false, alt: "2000")[$2000$], the population was #math.equation(block: false, alt: "1 , 545 , 387")[$1 , 545 , 387$]. Write a formula for the population of Riverside County. (You do not know the value of the growth factor, #math.equation(block: false, alt: "b")[$b$], yet.) + In #math.equation(block: false, alt: "2004")[$2004$], the population had grown to #math.equation(block: false, alt: "1 , 871 , 950")[$1 , 871 , 950$]. Find the growth factor and the percent rate of growth, rounded to the nearest tenth of a percent. + Estimate the population of Riverside County in #math.equation(block: false, alt: "2010")[$2010$]. + #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals 1 , 545 , 387 b to the power t")[$P ( t ) = 1 , 545 , 387 b^(t)$] + Growth factor #math.equation(block: false, alt: "1.049")[$1.049$]; Percent rate of growth #math.equation(block: false, alt: "4.9 %")[$4.9 upright(%)$] + #math.equation(block: false, alt: "2 , 493 , 401")[$2 , 493 , 401$] + In #math.equation(block: false, alt: "2006")[$2006$], a new Ford Focus cost #math.equation(block: false, alt: "$ 15 , 574")[$upright(\$) 15 , 574$]. The value of a Focus decreases exponentially over time. Write a formula for the value of a Focus. (You do not know the value of the decay factor, #math.equation(block: false, alt: "b")[$b$], yet.) + A #math.equation(block: false, alt: "2")[$2$]-year old Focus cost #math.equation(block: false, alt: "$ 11 , 788")[$upright(\$) 11 , 788$]. Find the decay factor and the percent rate of depreciation, rounded to the nearest tenth of a percent. + About how much would a #math.equation(block: false, alt: "4")[$4$]-year old Focus cost? In the 1940s, David Lack undertook a study of the European robin. He tagged #math.equation(block: false, alt: "130")[$130$] one-year-old robins and found that on average #math.equation(block: false, alt: "35.6 %")[$35.6 upright(%)$] of the birds survived each year. (Source: Burton, 1998) + According to the data, how many robins would have originally hatched to produce #math.equation(block: false, alt: "130")[$130$] one-year-olds? + Write a formula for the number of the original robins still alive after #math.equation(block: false, alt: "t")[$t$] years. + Graph your function. + One of the original robins actually survived for #math.equation(block: false, alt: "9")[$9$] years. How many robins does the model predict will survive for #math.equation(block: false, alt: "9")[$9$] years? + #math.equation(block: false, alt: "365")[$365$] + #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals 365 open parenthesis 0.356 close parenthesis to the power t")[$N ( t ) = 365 ( 0.356 )^(t)$] + #figure(figph[decay], alt: "decay", caption: none) + #math.equation(block: false, alt: "0.03")[$0.03$]. (Therefore, none) Many insects grow by discrete amounts each time they shed their exoskeletons. Dyar's rule says that the size of the insect increases by a constant ratio at each stage. (Source: Burton, 1998) + Dyar measured the width of the head of a caterpillar of a swallowtail butterfly at each stage. The caterpillar's head was initially approximately #math.equation(block: false, alt: "42")[$42$] millimeters wide, and #math.equation(block: false, alt: "63.84")[$63.84$] millimeters wide after its first stage. Find the growth ratio. + Write a formula for the width of the caterpillar's head at the #math.equation(block: false, alt: "n")[$n$]th stage. + Graph your function. + What head width does the model predict after #math.equation(block: false, alt: "5")[$5$] stages? For Problems 45–54, + Each table describes exponential growth or decay. Find the growth or decay factor. + Complete the table. Round values to two decimal places if necessary. #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]]), [#math.equation(block: false, alt: "P")[$P$]], [#math.equation(block: false, alt: "8")[$" " " " 8 " " " "$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "18")[$18$]], [$#hide($000$)$], [$#hide($000$)$], )) The growth factor is #math.equation(block: false, alt: "1.5")[$1.5$]. #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]]), [#math.equation(block: false, alt: "P")[$P$]], [#math.equation(block: false, alt: "8")[$" " 8 " "$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "18")[$18$]], [#math.equation(block: false, alt: "27")[$27$]], [#math.equation(block: false, alt: "40.5")[$40.5$]], )) #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]]), [#math.equation(block: false, alt: "P")[$P$]], [#math.equation(block: false, alt: "4")[$" " " " 4 " " " "$]], [#math.equation(block: false, alt: "5")[$" " " " 5 " " " "$]], [#math.equation(block: false, alt: "6.25")[$6.25$]], [$#hide($000$)$], [$#hide($000$)$], )) #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]]), [#math.equation(block: false, alt: "Q")[$Q$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "24")[$24$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) The growth factor is #math.equation(block: false, alt: "1.2")[$1.2$]. #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]]), [#math.equation(block: false, alt: "Q")[$Q$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "24")[$24$]], [#math.equation(block: false, alt: "28.8")[$28.8$]], [#math.equation(block: false, alt: "34.56")[$34.56$]], [#math.equation(block: false, alt: "41.47")[$41.47$]], )) #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]]), [#math.equation(block: false, alt: "Q")[$Q$]], [#math.equation(block: false, alt: "100")[$100$]], [#math.equation(block: false, alt: "105")[$105$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "w")[$w$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]]), [#math.equation(block: false, alt: "N")[$N$]], [#math.equation(block: false, alt: "120")[$120$]], [#math.equation(block: false, alt: "96")[$96$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) The decay factor is #math.equation(block: false, alt: "0.8")[$0.8$]. #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "w")[$w$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]]), [#math.equation(block: false, alt: "N")[$N$]], [#math.equation(block: false, alt: "120")[$120$]], [#math.equation(block: false, alt: "96")[$96$]], [#math.equation(block: false, alt: "76.8")[$76.8$]], [#math.equation(block: false, alt: "61.44")[$61.44$]], [#math.equation(block: false, alt: "49.15")[$49.15$]], )) #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "w")[$w$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]]), [#math.equation(block: false, alt: "N")[$N$]], [#math.equation(block: false, alt: "640")[$640$]], [#math.equation(block: false, alt: "480")[$480$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]]), [#math.equation(block: false, alt: "C")[$C$]], [#math.equation(block: false, alt: "10")[$10$]], [$#hide($000$)$], [#math.equation(block: false, alt: "6.4")[$6.4$]], [$#hide($000$)$], [$#hide($000$)$], )) The decay factor is #math.equation(block: false, alt: "0.8")[$0.8$]. #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]]), [#math.equation(block: false, alt: "C")[$C$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "6.4")[$6.4$]], [#math.equation(block: false, alt: "5.12")[$5.12$]], [#math.equation(block: false, alt: "4.10")[$4.10$]], )) #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]]), [#math.equation(block: false, alt: "C")[$C$]], [#math.equation(block: false, alt: "20")[$20$]], [$#hide($000$)$], [$#hide($000$)$], [#math.equation(block: false, alt: "2.5")[$2.5$]], [$#hide($000$)$], )) #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]]), [#math.equation(block: false, alt: "B")[$B$]], [#math.equation(block: false, alt: "200")[$200$]], [$#hide($000$)$], [$#hide($000$)$], [#math.equation(block: false, alt: "266.2")[$266.2$]], [$#hide($000$)$], )) The growth factor is #math.equation(block: false, alt: "1.1")[$1.1$]. #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]]), [#math.equation(block: false, alt: "B")[$B$]], [#math.equation(block: false, alt: "200")[$200$]], [#math.equation(block: false, alt: "220")[$220$]], [#math.equation(block: false, alt: "242")[$242$]], [#math.equation(block: false, alt: "266.2")[$266.2$]], [#math.equation(block: false, alt: "292.82")[$292.82$]], )) #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]]), [#math.equation(block: false, alt: "B")[$B$]], [#math.equation(block: false, alt: "40")[$40$]], [$#hide($000$)$], [#math.equation(block: false, alt: "62.5")[$62.5$]], [$#hide($000$)$], [$#hide($000$)$], )) Each graph in Problems 55–58 represents exponential growth or decay. + Find the initial value and the growth or decay factor. + Write a formula for the function. #figure(figph[growth], alt: "growth", caption: none) + Initial value #math.equation(block: false, alt: "4")[$4$], growth factor #math.equation(block: false, alt: "2 to the power 1 / 3")[$2^(1 / 3)$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 4 times 2 to the power x / 3")[$f ( x ) = 4 ⋅ 2^(x / 3)$] #figure(figph[growth], alt: "growth", caption: none) #figure(figph[decay], alt: "decay", caption: none) + Initial value #math.equation(block: false, alt: "80")[$80$], decay factor #math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 80 times open parenthesis the fraction 1 over 2 close parenthesis to the power x")[$f ( x ) = 80 ⋅ attach(( display(frac(1, 2)) ), t: x)$] #figure(figph[decay], alt: "decay", caption: none) If #math.equation(block: false, alt: "8 %")[$8 upright(%)$] of the air leaks out of Brian's bicycle tire every day, what percent of the air will be left after #math.equation(block: false, alt: "2")[$2$] days? After a week? #math.equation(block: true, alt: "84.6 %")[$84.6 upright(%)$], #math.equation(block: true, alt: "55.8 %")[$55.8 upright(%)$] If housing prices are increasing by #math.equation(block: false, alt: "15 %")[$15 upright(%)$] per year, by what percent will they increase in #math.equation(block: false, alt: "2")[$2$] years? In #math.equation(block: false, alt: "3")[$3$] years? Francine says that if a population grew by #math.equation(block: false, alt: "48 %")[$48 upright(%)$] in #math.equation(block: false, alt: "6")[$6$] years, then it grew by #math.equation(block: false, alt: "8 %")[$8 upright(%)$] per year. Is she correct? Either justify or correct her calculation. No, an increase of #math.equation(block: false, alt: "48 %")[$48 upright(%)$] in #math.equation(block: false, alt: "6")[$6$] years corresponds to a growth factor of #math.equation(block: false, alt: "1.48 to the power 1 / 6 approximately equals 1.0675")[$1.48^(1 / 6) ≈ 1.0675$], or an annual growth rate of about #math.equation(block: false, alt: "6.75 %")[$6.75 upright(%)$]. Delbert says that if a population decreased by #math.equation(block: false, alt: "60 %")[$60 upright(%)$] in 5 years, then it decreased by #math.equation(block: false, alt: "12 %")[$12 upright(%)$] per year. Is he correct? Either justify or correct his calculation. In Problems 63–66, assume that each population grows exponentially with constant annual percent increase, #math.equation(block: false, alt: "r")[$r$]. + The population of the state of Texas was #math.equation(block: false, alt: "16 , 986 , 335")[$16 , 986 , 335$] in #math.equation(block: false, alt: "1990")[$1990$]. Write a formula in terms of #math.equation(block: false, alt: "r")[$r$] for the population of Texas #math.equation(block: false, alt: "t")[$t$] years later. + In #math.equation(block: false, alt: "2000")[$2000$], the population was #math.equation(block: false, alt: "20 , 851 , 820")[$20 , 851 , 820$]. Write an equation and solve for #math.equation(block: false, alt: "r")[$r$]. What was the annual percent increase to the nearest hundredth of a percent? + #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals 16 , 986 , 335 open parenthesis 1 plus r close parenthesis to the power t")[$P ( t ) = 16 , 986 , 335 ( 1 + r )^(t)$] + #math.equation(block: false, alt: "2.07 %")[$2.07 upright(%)$] + The population of the state of Florida was #math.equation(block: false, alt: "12 , 937 , 926")[$12 , 937 , 926$] in #math.equation(block: false, alt: "1990")[$1990$]. Write a formula in terms of #math.equation(block: false, alt: "r")[$r$] for the population of Florida #math.equation(block: false, alt: "t")[$t$] years later. + In #math.equation(block: false, alt: "2000")[$2000$], the population was #math.equation(block: false, alt: "15 , 982 , 378")[$15 , 982 , 378$]. Write an equation and solve for #math.equation(block: false, alt: "r")[$r$]. What was the annual percent increase to the nearest hundredth of a percent? + The population of Rainville was #math.equation(block: false, alt: "10 , 000")[$10 , 000$] in #math.equation(block: false, alt: "1990")[$1990$] and doubled in #math.equation(block: false, alt: "20")[$20$] years. What was the annual percent increase to the nearest hundredth percent? + The population of Elmira was #math.equation(block: false, alt: "350 , 000")[$350 , 000$] in #math.equation(block: false, alt: "1990")[$1990$] and doubled in #math.equation(block: false, alt: "20")[$20$] years. What was the annual percent increase to the nearest hundredth of a percent? + If a population doubles in #math.equation(block: false, alt: "20")[$20$] years, does the percent increase depend on the size of the original population? + The population of Grayling doubled in #math.equation(block: false, alt: "20")[$20$] years. What was the annual percent increase to the nearest hundredth of a percent? + #math.equation(block: false, alt: "3.53 %")[$3.53 upright(%)$] + #math.equation(block: false, alt: "3.53 %")[$3.53 upright(%)$] + No + #math.equation(block: false, alt: "3.53 %")[$3.53 upright(%)$] + The population of Boomtown was #math.equation(block: false, alt: "300")[$300$] in #math.equation(block: false, alt: "1908")[$1908$] and tripled in #math.equation(block: false, alt: "7")[$7$] years. What was the annual percent increase to the nearest hundredth of a percent? + The population of Fairview was #math.equation(block: false, alt: "15 , 000")[$15 , 000$] in #math.equation(block: false, alt: "1962")[$1962$] and tripled in #math.equation(block: false, alt: "7")[$7$] years. What was the annual percent increase to the nearest hundredth of a percent? + If a population triples in #math.equation(block: false, alt: "7")[$7$] years, does the percent increase depend on the size of the original population? + The population of Pleasant Lake tripled in #math.equation(block: false, alt: "7")[$7$] years. What was the annual percent increase to the nearest hundredth of a percent? A researcher starts 2 populations of fruit flies of different species, each with #math.equation(block: false, alt: "30")[$30$] flies. Species A increases by #math.equation(block: false, alt: "30 %")[$30 upright(%)$] in #math.equation(block: false, alt: "6")[$6$] days and species B increases by #math.equation(block: false, alt: "20 %")[$20 upright(%)$] in #math.equation(block: false, alt: "4")[$4$] days. + What was the population of species A after #math.equation(block: false, alt: "6")[$6$] days? Find the daily growth factor for species A. + What was the population of species B after #math.equation(block: false, alt: "4")[$4$] days? Find the daily growth factor for species B. + Which species multiplies more rapidly? + #math.equation(block: false, alt: "39")[$39$]; #math.equation(block: false, alt: "1.045")[$1.045$] + #math.equation(block: false, alt: "35")[$35$]; #math.equation(block: false, alt: "1.047")[$1.047$] + Species B A biologist isolates two strains of a particular virus and monitors the growth of each, starting with samples of #math.equation(block: false, alt: "0.01")[$0.01$] gram. Strain A increases by #math.equation(block: false, alt: "10 %")[$10 upright(%)$] in #math.equation(block: false, alt: "8")[$8$] hours and strain B increases by #math.equation(block: false, alt: "12 %")[$12 upright(%)$] in #math.equation(block: false, alt: "9")[$9$] hours. + How much did the sample of strain A weigh after #math.equation(block: false, alt: "8")[$8$] hours? What was its hourly growth factor? + How much did the sample of strain B weigh after #math.equation(block: false, alt: "9")[$9$] hours? What was its hourly growth factor? + Which strain of virus grows more rapidly? In Problems 69–72, we compare linear and exponential growth. At a large university #math.equation(block: false, alt: "3")[$3$] students start a rumor that final exams have been canceled. After #math.equation(block: false, alt: "2")[$2$] hours, #math.equation(block: false, alt: "6")[$6$] students (including the first #math.equation(block: false, alt: "3")[$3$]) have heard the rumor. + Assuming that the rumor grows linearly, complete the table below for #math.equation(block: false, alt: "L open parenthesis t close parenthesis")[$L ( t )$], the number of students who have heard the rumor after #math.equation(block: false, alt: "t")[$t$] hours. Then write a formula for the function #math.equation(block: false, alt: "L open parenthesis t close parenthesis")[$L ( t )$]. Graph the function.#figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "8")[$8$]]), [#math.equation(block: false, alt: "L open parenthesis t close parenthesis")[$L ( t )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Complete the table below, assuming that the rumor grows exponentially. Write a formula for the function #math.equation(block: false, alt: "E open parenthesis t close parenthesis")[$E ( t )$] and graph it on the same set of axes with #math.equation(block: false, alt: "L open parenthesis t close parenthesis")[$L ( t )$].#figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "8")[$8$]]), [#math.equation(block: false, alt: "E open parenthesis t close parenthesis")[$E ( t )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "8")[$8$]]), [#math.equation(block: false, alt: "L open parenthesis t close parenthesis")[$L ( t )$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "15")[$15$]], )) #linebreak() #math.equation(block: false, alt: "L open parenthesis t close parenthesis equals 3 plus 1.5 t")[$L ( t ) = 3 + 1.5 t$]#figure(figph[linear and exponential], alt: "linear and exponential", caption: none) + #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "8")[$8$]]), [#math.equation(block: false, alt: "E open parenthesis t close parenthesis")[$E ( t )$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "24")[$24$]], [#math.equation(block: false, alt: "48")[$48$]], )) #linebreak() #math.equation(block: false, alt: "E open parenthesis t close parenthesis equals 3 times 2 to the power t / 2")[$E ( t ) = 3 ⋅ 2^(t / 2)$] Over the weekend the Midland Infirmary identifies four cases of Asian flu. Three days later it has treated a total of ten cases. + Assuming that the number of flu cases grows linearly, complete the table below for #math.equation(block: false, alt: "L open parenthesis t close parenthesis")[$L ( t )$], the number of people infected after #math.equation(block: false, alt: "t")[$t$] days. Then write a formula for the function #math.equation(block: false, alt: "L open parenthesis t close parenthesis")[$L ( t )$]. Graph the function.#figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "12")[$12$]]), [#math.equation(block: false, alt: "L open parenthesis t close parenthesis")[$L ( t )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Complete the table below, assuming that the flu grows exponentially. Write a formula for the function #math.equation(block: false, alt: "E open parenthesis t close parenthesis")[$E ( t )$] and graph it on the same set of axes with #math.equation(block: false, alt: "L open parenthesis t close parenthesis")[$L ( t )$].#figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "12")[$12$]]), [#math.equation(block: false, alt: "E open parenthesis t close parenthesis")[$E ( t )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) The world’s population of tigers declined from #math.equation(block: false, alt: "10 , 400")[$10 , 400$] in #math.equation(block: false, alt: "1980")[$1980$] to #math.equation(block: false, alt: "6000")[$6000$] in #math.equation(block: false, alt: "1998")[$1998$]. + If the population declined linearly, what was its annual rate of decrease? + If the population declined exponentially, what was its annual decay factor? What was its annual percent decrease? + Predict the number of tigers in #math.equation(block: false, alt: "2010")[$2010$] under each assumption, linear or exponential decline. + #math.equation(block: false, alt: "244")[$244$] tigers per year + #math.equation(block: false, alt: "0.97")[$0.97$]; #math.equation(block: false, alt: "3 %")[$3 upright(%)$] + Linear: #math.equation(block: false, alt: "3067")[$3067$]; Exponential: #math.equation(block: false, alt: "4170")[$4170$] In 2003, the Center for Biological Diversity filed a lawsuit against the federal government for failing to protect Alaskan sea otters. The population of sea otters, which numbered between 150,000 and 300,000 before hunting began in 1741, declined from about 20,000 in 1992 to 6000 in 2000. (Source: Center for Biological Diversity) + If the population declined linearly after 1992, what was its annual rate of change in population? + If the population declined exponentially after 1992, what was its annual decay factor? + Predict the number of sea otters in 2010 under each assumption, linear or exponential decline