#set document(title: "5.3 The Natural Base", 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")) == 5.3#h(0.6em)The Natural Base We have looked at logarithms with various bases, and in particular we studied the common or base 10 logarithms, which often appear in applications. There is another base for logarithms and exponential functions that is often used in applications. This base is an irrational number called #math.equation(block: false, alt: "e")[$e$], where #math.equation(block: true, alt: "e approximately equals 2.71828182845")[$bold(italic(e)) ≈ 2.71828182845$] The number #math.equation(block: false, alt: "e")[$e$] is essential for many advanced topics, and it is often called the #strong[natural base]. It may seem strange to use an irrational number as the base for exponential functions, but just as the irrational number #math.equation(block: false, alt: "π")[$π$] arises naturally in geometry, so does #math.equation(block: false, alt: "e")[$e$] arise in calculus and its applications. At the end of this section we'll look at a specific case of how base #math.equation(block: false, alt: "e")[$e$] occurs, and how its use is connected to the ideas and techniques of calculus. === The Natural Exponential Function The #strong[natural exponential function] is the function #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals e to the power x")[$f ( x ) = e^(x)$] Values for #math.equation(block: false, alt: "e to the power x")[$e^(x)$] can be obtained with a calculator using the #math.equation(block: false, alt: "e to the power x")[$#box(stroke: 0.6pt, inset: 3pt, outset: 0pt)[$display(e^(x))$]$] key ( 2nd LN on most calculators). For example, you can evaluate #math.equation(block: false, alt: "e to the power 1")[$e^(1)$] by pressing 2nd LN #math.equation(block: false, alt: "1")[$1$] to confirm the value of #math.equation(block: false, alt: "e")[$e$] given above. (We'll explain why we use the 2nd LN key a little later.) Try a few more calculations to become familiar with base #math.equation(block: false, alt: "e")[$e$]. Use your calculator to evaluate the following powers. + #math.equation(block: false, alt: "e squared approximately equals")[$e^(2) ≈$]\_\_\_\_\_ + #math.equation(block: false, alt: "e to the power 3.5 approximately equals")[$e^(3.5) ≈$]\_\_\_\_\_ + #math.equation(block: false, alt: "e to the power minus 0.5 approximately equals")[$e^(− 0.5) ≈$]\_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "e squared approximately equals 7.389")[$e^(2) ≈ 7.389$] + #math.equation(block: false, alt: "e to the power 3.5 approximately equals 33.115")[$e^(3.5) ≈ 33.115$] + #math.equation(block: false, alt: "e to the power minus 0.5 approximately equals 0.6065")[$e^(− 0.5) ≈ 0.6065$] ] Use your calculator to evaluate the following powers. Round to four decimal places. + #math.equation(block: false, alt: "e squared")[$e^(2)$] + #math.equation(block: false, alt: "e to the power 3.5")[$e^(3.5)$] + #math.equation(block: false, alt: "e to the power minus 0.5")[$e^(− 0.5)$] #solutionbox[ + #math.equation(block: false, alt: "e squared approximately equals 7.3891")[$e^(2) ≈ 7.3891$] + #math.equation(block: false, alt: "e to the power 3.5 approximately equals 33.1155")[$e^(3.5) ≈ 33.1155$] + #math.equation(block: false, alt: "e to the power minus 0.5 approximately equals 0.6065")[$e^(− 0.5) ≈ 0.6065$] ] What about the graph of this new exponential function? Because #math.equation(block: false, alt: "e")[$e$] is a number between #math.equation(block: false, alt: "2")[$2$] and #math.equation(block: false, alt: "3")[$3$], the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals e to the power x")[$" " f ( x ) = e^(x) " "$] lies between the graphs of #math.equation(block: false, alt: "y equals 2 to the power x")[$y = 2^(x)$] and #math.equation(block: false, alt: "y equals 3 to the power x")[$y = 3^(x)$]. Compare the tables of values and the graphs of the three functions below. For example, note that for #math.equation(block: false, alt: "x equals 2")[$x = 2$], the value of #math.equation(block: false, alt: "e squared equals 7.389")[$e^(2) = 7.389$] is between #math.equation(block: false, alt: "2 squared equals 4")[$2^(2) = 4$] and #math.equation(block: false, alt: "3 squared equals 9")[$3^(2) = 9$]. You can verify the table and graphs on your calculator. #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y equals 2 to the power x")[$y = 2^(x)$]], [#math.equation(block: false, alt: "y equals e to the power x")[$y = e^(x)$]], [#math.equation(block: false, alt: "y equals 3 to the power x")[$y = 3^(x)$]]), [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "0.125")[$0.125$]], [#math.equation(block: false, alt: "0.050")[$0.050$]], [#math.equation(block: false, alt: "0.037")[$0.037$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "0.250")[$0.250$]], [#math.equation(block: false, alt: "0.135")[$0.135$]], [#math.equation(block: false, alt: "0.111")[$0.111$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0.500")[$0.500$]], [#math.equation(block: false, alt: "0.368")[$0.368$]], [#math.equation(block: false, alt: "0.333")[$0.333$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2.718")[$2.718$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "7.389")[$7.389$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "20.086")[$20.086$]], [#math.equation(block: false, alt: "27")[$27$]], )) #figure(figph[graph of 3 exponential functions], alt: "graph of 3 exponential functions", caption: none) The value of #math.equation(block: false, alt: "e squared")[$e^(2)$] is closest to \_\_\_\_\_ #solutionbox[ 7 ] + The value of #math.equation(block: false, alt: "e squared")[$e^(2)$] is closest to which of these?+ #math.equation(block: false, alt: "3")[$3$] + #math.equation(block: false, alt: "5")[$5$] + #math.equation(block: false, alt: "7")[$7$] + #math.equation(block: false, alt: "9")[$9$] + The value of #math.equation(block: false, alt: "e to the power minus 1")[$e^(− 1)$] is closest to which of these?+ #math.equation(block: false, alt: "minus 3")[$− 3$] + #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$] + #math.equation(block: false, alt: "the fraction 1 over 3")[$display(frac(1, 3))$] + #math.equation(block: false, alt: "2")[$2$] Variations on the natural exponential function occur in many disciplines. For example, the graph in the figure below is called a "bell curve." It is the graph of the #strong[normal distribution] in statistics. #figure(figph[bell curve of male height], alt: "bell curve of male height", caption: none) #examplebox("Example 1")[][ The normal distribution above shows the heights of men in the US. The average or mean height for men is 70 inches, and the formula for the graph is #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 0.13 e to the power minus open parenthesis x minus 70 close parenthesis squared / 18")[$f ( x ) = 0.13 " " e^(− ( x − 70 )^(2) / 18)$] The probability that a given man has a height between two values is the area under the curve between those values. Evaluate the function for #math.equation(block: false, alt: "x equals 70")[$x = 70$] and for #math.equation(block: false, alt: "x equals 67")[$x = 67$]. #solutionbox[ To evaluate the function we follow the order of operations. For #math.equation(block: false, alt: "x equals 70")[$x = 70$] we have #math.equation(block: true, alt: "f open parenthesis 70 close parenthesis, equals 0.13 e to the power minus open parenthesis 70 minus 70 close parenthesis squared / 18, Simplify the exponent first.; equals 0.13 e to the power 0 equals 0.13")[$f ( 70 ) & = 0.13 " " e^(− ( 70 − 70 )^(2) / 18) & & "Simplify the exponent first." \ & = 0.13 " " e^(0) = 0.13$] For #math.equation(block: false, alt: "x equals 67")[$x = 67$] we have #math.equation(block: true, alt: "f open parenthesis 67 close parenthesis, equals 0.13 e to the power minus open parenthesis 67 minus 70 close parenthesis squared / 18, minus open parenthesis 67 minus 70 close parenthesis squared / 18 equals minus 0.5; equals 0.13 e to the power minus 0.5 equals 0.079")[$f ( 67 ) & = 0.13 " " e^(− ( 67 − 70 )^(2) / 18) & & − ( 67 − 70 )^(2) / 18 = − 0.5 \ & = 0.13 " " e^(− 0.5) = 0.079$] ] ] This next graph is an example of a #strong[logistic function], which models population growth with an upper bound. #figure(figph[logistic graph of covid], alt: "logistic graph of covid", caption: none) The logistic function shown above models the spread of Covid in China during the 2020 epidemic. It gives the number of infections #math.equation(block: false, alt: "N open parenthesis t close parenthesis")[$N ( t )$], in thousands, reported #math.equation(block: false, alt: "t")[$t$] days after January 21, 2020. The equation for this model is #math.equation(block: true, alt: "N open parenthesis t close parenthesis equals the fraction 83.5 over 1 plus 267 e to the power minus 0.214 t")[$N ( t ) = display(frac(83.5, 1 + 267 " " e^(− 0.214 t)))$] According to the model, how many cases of Covid were reported on February 20 (day #math.equation(block: false, alt: "t equals 30")[$t = 30$])? #solutionbox[ 58,194 cases ] #examplebox("Example 2")[][ Use technology to graph each function. How does each graph differ from the graph of #math.equation(block: false, alt: "y equals e to the power x")[$y = e^(x)$]? + #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals e to the power x plus 2")[$g ( x ) = e^(x + 2)$] + #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals e to the power x plus 2")[$h ( x ) = e^(x) + 2$] #solutionbox[ #figure(figph[translations of the exponential], alt: "translations of the exponential", caption: none) The graph of #math.equation(block: false, alt: "g")[$g$] is shifted #math.equation(block: false, alt: "2")[$2$] units to the left of #math.equation(block: false, alt: "y equals e to the power x")[$y = e^(x)$]. The graph of #math.equation(block: false, alt: "h")[$h$] is shifted #math.equation(block: false, alt: "2")[$2$] units up from #math.equation(block: false, alt: "y equals e to the power x")[$y = e^(x)$]. The graphs are shown above. ] ] === The Natural Logarithmic Function Recall that each exponential function with base #math.equation(block: false, alt: "b")[$b$] has an inverse function, the logarithmic function with the same base. For example, the function #math.equation(block: false, alt: "y equals log sub 2 ⁡ open parenthesis x close parenthesis")[$" " y = log_(2) ( x ) " "$] is the inverse of the function #math.equation(block: false, alt: "y equals 2 to the power x")[$" " y = 2^(x)$]. It gives the exponent needed on 2 to give #math.equation(block: false, alt: "x")[$x$], so that, for instance, #math.equation(block: false, alt: "log sub 2 ⁡ open parenthesis 16 close parenthesis equals 4")[$" " log_(2) ( 16 ) = 4 " "$], because #math.equation(block: false, alt: "2 to the power 4 equals 16")[$" " 2^(4) = 16$]. The base #math.equation(block: false, alt: "e")[$e$] logarithm of a number #math.equation(block: false, alt: "x")[$x$], or #math.equation(block: false, alt: "log sub e ⁡ open parenthesis x close parenthesis")[$log_(e) ( x )$], is called the #strong[natural logarithm] of #math.equation(block: false, alt: "x")[$x$] and is denoted by #math.equation(block: false, alt: "ln ⁡ open parenthesis x close parenthesis")[$ln ( x )$] or #math.equation(block: false, alt: "ln ⁡ x")[$ln x$]. (Why “ln” and not “nl”? The natural logarithm is denoted by “ln” because it stands for “logarithmus naturalis,” which is the Latin for “natural logarithm.”) Here is its official definition. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[The Natural Logarithm] The natural logarithm is the logarithm base #math.equation(block: false, alt: "e")[$e$]. #math.equation(block: true, alt: "ln ⁡ open parenthesis x close parenthesis equals log sub e ⁡ open parenthesis x close parenthesis , x greater than 0")[$ln ( x ) = log_(e) ( x ) , " " " " " " " " x > 0$] ] The natural logarithm of #math.equation(block: false, alt: "x")[$x$] is the exponent to which #math.equation(block: false, alt: "e")[$e$] must be raised to produce #math.equation(block: false, alt: "x")[$x$]. For example, the natural logarithm of #math.equation(block: false, alt: "10")[$10$], or #math.equation(block: false, alt: "ln ⁡ open parenthesis 10 close parenthesis")[$ln ( 10 )$], is the solution of the equation #math.equation(block: true, alt: "e to the power y equals 10")[$e^(y) = 10$] You can verify on your calculator that #math.equation(block: true, alt: "e to the power 2.3 approximately equals 10 or ln ⁡ open parenthesis 10 close parenthesis approximately equals 2.3")[$e^(2.3) " " " " ≈ 10 " or " " " " " ln ( 10 ) ≈ 2.3$] As is the case with exponential and log functions with other bases, the #strong[natural log function], #math.equation(block: false, alt: "y equals ln ⁡ open parenthesis x close parenthesis")[$y = ln ( x )$], and the natural exponential function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals e to the power x")[$f ( x ) = e^(x)$], “undo” each other, so they are inverse functions. (This is why many calculators use 2nd LN to indicate #math.equation(block: false, alt: "e to the power x")[$e^(x)$].) Use your calculator to evaluate each logarithm. Round your answers to four decimal places. + #math.equation(block: false, alt: "ln ⁡ 100 approximately equals")[$ln 100 ≈$]\_\_\_\_\_ + #math.equation(block: false, alt: "ln ⁡ 0.01 approximately equals")[$ln 0.01 ≈$]\_\_\_\_\_ + #math.equation(block: false, alt: "ln ⁡ e cubed approximately equals")[$ln e^(3) ≈$]\_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "ln ⁡ 100 approximately equals 4.6052")[$ln 100 ≈ 4.6052$] + #math.equation(block: false, alt: "ln ⁡ 0.01 approximately equals minus 4.6052")[$ln 0.01 ≈ − 4.6052$] + #math.equation(block: false, alt: "ln ⁡ e cubed equals 3")[$ln e^(3) = 3$] ] Use your calculator to evaluate each logarithm. Round your answers to four decimal places. + #math.equation(block: false, alt: "ln ⁡ open parenthesis 10 close parenthesis")[$ln ( 10 )$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis 0.1 close parenthesis")[$ln ( 0.1 )$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis 100 close parenthesis")[$ln ( 100 )$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis 0.01 close parenthesis")[$ln ( 0.01 )$] #solutionbox[ + #math.equation(block: false, alt: "ln ⁡ open parenthesis 10 close parenthesis approximately equals 2.3026")[$ln ( 10 ) ≈ 2.3026$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis 0.1 close parenthesis approximately equals minus 2.3026")[$ln ( 0.1 ) ≈ − 2.3026$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis 100 close parenthesis approximately equals 4.6052")[$ln ( 100 ) ≈ 4.6052$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis 0.01 close parenthesis approximately equals minus 4.6052")[$ln ( 0.01 ) ≈ − 4.6052$] ] As usual, we can gain a better understanding of a new function by looking at its graph. #examplebox("Example 3")[][ + Graph #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals e to the power x")[$" " f ( x ) = e^(x) " "$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals ln ⁡ open parenthesis x close parenthesis")[$" " g ( x ) = ln ( x ) " "$] on the same grid. + Give the domain and range of the natural log function. #solutionbox[ Earlier we made a table of values to graph the fuction #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals e to the power x")[$" " f ( x ) = e^(x)$], and we can make a table of values for #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals ln ⁡ open parenthesis x close parenthesis")[$g ( x ) = ln ( x )$] by interchanging the columns in that table. Plotting the points gives us the graph below. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y equals ln ⁡ open parenthesis x close parenthesis")[$y = ln ( x )$]]), [#math.equation(block: false, alt: "0.050")[$0.050$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "0.135")[$0.135$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "0.368")[$0.368$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "2.718")[$2.718$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "7.389")[$7.389$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "20.086")[$20.086$]], [#math.equation(block: false, alt: "3")[$3$]], )) #figure(figph[graphs of ln and natural exponential], alt: "graphs of ln and natural exponential", caption: none) You can see that the graph of #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals ln ⁡ open parenthesis x close parenthesis")[$" " g ( x ) = ln ( x ) " "$] is the reflection of the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals e to the power x")[$" " f ( x ) = e^(x) " "$] about the line #math.equation(block: false, alt: "y equals x")[$y = x$]. In particular, notice that while the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals e to the power x")[$f ( x ) = e^(x)$] has no points with negative #math.equation(block: false, alt: "y")[$y$]-values, the graph of #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals ln ⁡ open parenthesis x close parenthesis")[$" " g ( x ) = ln ( x ) " "$] has no points with negative #math.equation(block: false, alt: "x")[$x$]-values. Thus, the domain of the natural log function is the same as the range of #math.equation(block: false, alt: "y equals e to the power x")[$y = e^(x)$], or all positive numbers. The range of #math.equation(block: false, alt: "y equals ln ⁡ open parenthesis x close parenthesis")[$y = ln ( x )$] is the same as the domain of #math.equation(block: false, alt: "y equals e to the power x")[$y = e^(x)$], or all real numbers. ] ] From the graph of #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals ln ⁡ open parenthesis x close parenthesis")[$g ( x ) = ln ( x )$] you can make the following observations. - The natural log function has only positive numbers as input values. - The natural logs of negative numbers and zero are undefined. - The natural log of a number greater than 1 is positive, while the logs of numbers between 0 and 1 are negative. === Properties of the Natural Logarithm Natural logs obey the same conversion formulas that work for logs to other bases. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Conversion Formulas for Natural Logs] #math.equation(block: true, alt: "y equals ln ⁡ open parenthesis x close parenthesis if and only if e to the power y equals x")[$bold(italic(y)) = bold(ln) ( bold(italic(x)) ) " " " " " if and only if " " " " " bold(italic(e))^(bold(italic(y))) = bold(italic(x))$] ] The conversion formulas are just another way of saying the the the natural log function, #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals ln ⁡ open parenthesis x close parenthesis")[$g ( x ) = ln ( x )$], and the natural exponential function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals e to the power x")[$f ( x ) = e^(x)$], are inverse functions. In particular, #math.equation(block: true, alt: "ln ⁡ open parenthesis e close parenthesis, equals 1 because e to the power 1 equals e; ln ⁡ open parenthesis 1 close parenthesis, equals 0 because e to the power 0 equals 1")[$ln ( e ) & = 1 " because " e^(1) = e \ ln ( 1 ) & = 0 " because " e^(0) = 1$] Which of the following is equivalent to #math.equation(block: false, alt: "e to the power x equals k")[$e^(x) = k$] ? \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "ln ⁡ k equals x")[$ln k = x$] ] Which of the following is equivalent to #math.equation(block: false, alt: "e to the power x equals k")[$e^(x) = k$] ? + #math.equation(block: false, alt: "e to the power k equals x")[$e^(k) = x$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis e close parenthesis equals x")[$ln ( e ) = x$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis k close parenthesis equals x")[$ln ( k ) = x$] + #math.equation(block: false, alt: "k to the power x equals e")[$k^(x) = e$] We use natural logarithms in the same way that we use logs to other bases. The properties of logarithms that we studied in Properties of Logarithms also apply to logarithms base #math.equation(block: false, alt: "e")[$e$]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Properties of Natural Logarithms] If #math.equation(block: false, alt: "x , y greater than 0")[$x , y > 0$], then + #math.equation(block: false, alt: "ln ⁡ open parenthesis x y close parenthesis equals ln ⁡ open parenthesis x close parenthesis plus ln ⁡ open parenthesis y close parenthesis")[$ln ( x y ) = ln ( x ) + ln ( y )$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis the fraction x over y close parenthesis equals ln ⁡ open parenthesis x close parenthesis minus ln ⁡ open parenthesis y close parenthesis")[$ln ( display(frac(x, y)) ) = ln ( x ) − ln ( y )$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis x to the power k close parenthesis equals k ln ⁡ open parenthesis x close parenthesis")[$ln ( x^(k) ) = k ln ( x )$] ] Because the functions #math.equation(block: false, alt: "y equals e to the power x")[$" " y = e^(x) " "$] and #math.equation(block: false, alt: "y equals ln ⁡ x")[$" " y = ln x " "$] are inverse functions, the following properties are also true. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[The Natural log and #math.equation(block: false, alt: "e to the power x")[$e^(x)$]] #math.equation(block: true, alt: "ln ⁡ e to the power x equals x , for all x , and e to the power ln ⁡ open parenthesis x close parenthesis equals x , for x greater than 0")[$ln e^(x) = x , " " " " " for all " x , " " " " " " " " " " " and " " " " " " " e^(ln ( x )) = x , " " " " " for " x > 0$] ] #examplebox("Example 4")[][ Simplify each expression. + #math.equation(block: false, alt: "ln ⁡ open parenthesis e to the power 0.3 x close parenthesis")[$ln ( e^(0.3 x) )$] + #math.equation(block: false, alt: "e to the power 2 ln ⁡ open parenthesis x plus 3 close parenthesis")[$e^(2 ln ( x + 3 ))$] #solutionbox[ + The natural log is the log base #math.equation(block: false, alt: "e")[$e$], and hence the inverse of #math.equation(block: false, alt: "e to the power x")[$e^(x)$]. Therefore, #math.equation(block: true, alt: "ln ⁡ open parenthesis e to the power 0.3 x close parenthesis equals 0.3 x")[$ln ( e^(0.3 x) ) = 0.3 x$] + First, we simplify the exponent using the third property of logs to get #math.equation(block: true, alt: "2 ln ⁡ open parenthesis x plus 3 close parenthesis equals ln ⁡ open parenthesis open parenthesis x plus 3 close parenthesis squared close parenthesis")[$2 ln ( x + 3 ) = ln ( ( x + 3 )^(2) )$] Then #math.equation(block: false, alt: "e to the power 2 ln ⁡ open parenthesis x plus 3 close parenthesis equals e to the power ln ⁡ open parenthesis open parenthesis x plus 3 close parenthesis squared close parenthesis equals open parenthesis x plus 3 close parenthesis squared")[$e^(2 ln ( x + 3 )) = e^(ln ( ( x + 3 )^(2) )) = ( x + 3 )^(2)$]. ] ] Simplify each expression. Use "sqrt(x)" to get #math.equation(block: false, alt: "the square root of x")[$sqrt(x)$]. + #math.equation(block: false, alt: "e to the power open parenthesis ln ⁡ x close parenthesis / 2 equals")[$e^(( ln x ) / 2) =$]\_\_\_\_\_ + #math.equation(block: false, alt: "ln ⁡ open parenthesis the fraction 1 over e to the power 4 x close parenthesis equals")[$ln ( display(frac(1, e^(4 x))) ) =$]\_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "the square root of x")[$sqrt(x)$] + #math.equation(block: false, alt: "minus 4 x")[$− 4 x$] ] Simplify each expression. + #math.equation(block: false, alt: "e to the power open parenthesis ln ⁡ open parenthesis x close parenthesis / 2 close parenthesis")[$e^(( ln ( x ) / 2 ))$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis the fraction 1 over e to the power 4 x close parenthesis")[$ln ( display(frac(1, e^(4 x))) )$] #solutionbox[ + #math.equation(block: false, alt: "the square root of x")[$sqrt(x) " "$] or #math.equation(block: false, alt: "x to the power 1 / 2")[$" " x^(1 / 2)$] + #math.equation(block: false, alt: "minus 4 x")[$− 4 x$] ] Explain why #math.equation(block: false, alt: "ln ⁡ open parenthesis the fraction 1 over e cubed close parenthesis equals minus 3")[$ln ( display(frac(1, e^(3))) ) = − 3$]. \_\_\_\_\_ Explain why #math.equation(block: false, alt: "ln ⁡ open parenthesis the fraction 1 over e cubed close parenthesis equals minus 3")[$ln ( display(frac(1, e^(3))) ) = − 3$]. === Solving Equations We use the natural logarithm to solve exponential equations with base #math.equation(block: false, alt: "e")[$e$]. The techniques we've learned for solving other exponential equations also apply to equations with base #math.equation(block: false, alt: "e")[$e$]. #examplebox("Example 5")[][ Solve each equation for #math.equation(block: false, alt: "x")[$x$]. + #math.equation(block: false, alt: "e to the power x equals 0.24")[$e^(x) = 0.24$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis x close parenthesis equals 3.5")[$ln ( x ) = 3.5$] #solutionbox[ + We convert the equation to logarithmic form and evaluate using a calculator. #math.equation(block: true, alt: "x equals ln ⁡ open parenthesis 0.24 close parenthesis approximately equals minus 1.427")[$x = ln ( 0.24 ) ≈ − 1.427$] + We convert the equation to exponential form and evaluate. #math.equation(block: true, alt: "x equals e to the power 3.5 approximately equals 33.1155")[$x = e^(3.5) ≈ 33.1155$] ] ] Solve each equation. Round your answers to four decimal places. + #math.equation(block: false, alt: "ln ⁡ x equals minus 0.2")[$ln x = − 0.2$] #linebreak() #math.equation(block: false, alt: "x approximately equals")[$x ≈$]\_\_\_\_\_ + #math.equation(block: false, alt: "e to the power x equals 8")[$e^(x) = 8$] #linebreak() #math.equation(block: false, alt: "x approximately equals")[$x ≈$]\_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "0.8187")[$0.8187$] + #math.equation(block: false, alt: "2.0794")[$2.0794$] ] Solve each equation. Round your answers to four decimal places. + #math.equation(block: false, alt: "ln ⁡ open parenthesis x close parenthesis equals minus 0.2")[$ln ( x ) = − 0.2$] + #math.equation(block: false, alt: "e to the power x equals 8")[$e^(x) = 8$] #solutionbox[ + #math.equation(block: false, alt: "0.8187")[$0.8187$] + #math.equation(block: false, alt: "2.0794")[$2.0794$] ] Why is the equation #math.equation(block: false, alt: "e to the power x equals 6.5")[$e^(x) = 6.5$] easier to solve than #math.equation(block: false, alt: "8 to the power x equals 6.5")[$8^(x) = 6.5$] ? \_\_\_\_\_ #solutionbox[ There is a button for log base on the calculator, but not a button for log base 8. ] Which statement below explains why the equation #math.equation(block: false, alt: "e to the power x equals 6.5")[$e^(x) = 6.5$] is easier to solve than #math.equation(block: false, alt: "8 to the power x equals 6.5")[$8^(x) = 6.5$] ? + 8 is larger than 6.5. + #math.equation(block: false, alt: "k")[$k$] is a constant. + There is a button for log base #math.equation(block: false, alt: "e")[$e$] on the calculator, but not a button for log base 8. + Because #math.equation(block: false, alt: "e")[$e$] is an irrational number. To solve more complicated exponential equations, we isolate the power on one side of the equation before converting to logarithmic form. #examplebox("Example 6")[][ Solve #math.equation(block: false, alt: "200 equals 60 plus 20 e to the power 0.4 x")[$" " 200 = 60 + 20 e^(0.4 x)$] #solutionbox[ First, we isolate the power. #math.equation(block: true, alt: "200, equals 60 plus 20 e to the power 0.4 x, Subtract 60 from both sides.; 140, equals 20 e to the power 0.4 x, Divide both sides by 20.; 7, equals e to the power 0.4 x")[$200 & = 60 + 20 e^(0.4 x) & & "Subtract 60 from both sides." \ 140 & = 20 e^(0.4 x) & & "Divide both sides by 20." \ 7 & = e^(0.4 x)$] Then we convert the equation to logarithmic form. #math.equation(block: true, alt: "0.4 x, equals ln ⁡ open parenthesis 7 close parenthesis, Divide both sides by 0.4.; x, equals the fraction ln ⁡ open parenthesis 7 close parenthesis over 0.4")[$0.4 x & = ln ( 7 ) & & "Divide both sides by 0.4." \ x & = frac(ln ( 7 ), 0.4)$] Rounded to four decimal places, #math.equation(block: false, alt: "x approximately equals 4.8648")[$x ≈ 4.8648$]. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ We can also solve the equation in Example, #math.equation(block: true, alt: "7 equals e to the power 0.4 x")[$7 = e^(0.4 x)$] by taking the natural logarithm of both sides. This gives us #math.equation(block: true, alt: "ln ⁡ open parenthesis 7 close parenthesis, equals ln ⁡ e to the power 0.4 x, Simplify the right side.; ln ⁡ open parenthesis 7 close parenthesis, equals 0.4 x")[$ln ( 7 ) & = ln e^(0.4 x) & & "Simplify the right side." \ ln ( 7 ) & = 0.4 x$] because #math.equation(block: false, alt: "ln ⁡ open parenthesis e to the power a close parenthesis equals a")[$ln ( e^(a) ) = a$] for any number #math.equation(block: false, alt: "a")[$a$]. We then proceed with the solution as before. ] Solve #math.equation(block: true, alt: "80 minus 16 e to the power minus 0.2 x equals 70.3")[$80 − 16 e^(− 0.2 x) = 70.3$] #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "x equals minus 5 ln ⁡ open parenthesis the fraction 9.7 over 16 close parenthesis approximately equals 2.5023")[$x = − 5 ln ( display(frac(9.7, 16)) ) ≈ 2.5023$] ] Solve #math.equation(block: true, alt: "80 minus 16 e to the power minus 0.2 x equals 70.3")[$" " 80 − 16 e^(− 0.2 x) = 70.3$] #solutionbox[ Isolate the power, take the natural log of both sides, and solve as usual to find #math.equation(block: false, alt: "x equals minus 5 ln ⁡ open parenthesis the fraction 9.7 over 16 close parenthesis approximately equals 2.5023")[$x = − 5 ln ( display(frac(9.7, 16)) ) ≈ 2.5023$] ] #examplebox("Example 7")[][ The logistic function #math.equation(block: true, alt: "P equals the fraction 20 over 1 plus 5 e to the power minus 0.4 t")[$P = display(frac(20, 1 + 5 e^(− 0.4 t)))$] has its upper bound at #math.equation(block: false, alt: "P equals 20")[$" " P = 20$]. Find the value of #math.equation(block: false, alt: "t")[$t$] for which #math.equation(block: false, alt: "P equals 10")[$" " P = 10$]. #solutionbox[ We must solve the equation #math.equation(block: false, alt: "10 equals the fraction 20 over 1 plus 5 e to the power minus 0.4 t")[$" " 10 = display(frac(20, 1 + 5 e^(− 0.4 t))) " "$] for #math.equation(block: false, alt: "t")[$t$]. To begin, we multiply both sides of the equation by the denominator, #math.equation(block: false, alt: "1 plus 5 e to the power minus 0.4 t")[$1 + 5 e^(− 0.4 t)$], to get #math.equation(block: true, alt: "1 plus 5 e to the power minus 0.4 t equals the fraction 20 over 10 equals 2")[$1 + 5 e^(− 0.4 t) = display(frac(20, 10)) = 2$] Then we isolate the power, #math.equation(block: false, alt: "e to the power minus k t")[$e^(− k t)$], as follows: #math.equation(block: true, alt: "1 plus 5 e to the power minus 0.4 t, equals 2, Subtract 1 from both sides, then divide by 5.; e to the power minus 0.4 k t, equals 0.2")[$1 + 5 e^(− 0.4 t) & = 2 & & "Subtract 1 from both sides, then divide by 5." \ e^(− 0.4 k t) & = 0.2$] Next, we take the natural logarithm of both sides to get #math.equation(block: true, alt: "ln ⁡ open parenthesis e to the power minus 0.4 t close parenthesis equals ln ⁡ open parenthesis 0.2 close parenthesis")[$ln ( e^(− 0.4 t) ) = ln ( 0.2 )$] and recall that #math.equation(block: false, alt: "ln ⁡ open parenthesis e to the power x close parenthesis equals x")[$ln ( e^(x) ) = x$] to simplify the left side. #math.equation(block: true, alt: "minus 0.4 t equals ln ⁡ open parenthesis 0.2 close parenthesis")[$− 0.4 t = ln ( 0.2 )$] Finally, we divide both sides by #math.equation(block: false, alt: "minus 0.4")[$− 0.4$] to solve for #math.equation(block: false, alt: "t")[$t$]. #math.equation(block: true, alt: "t equals the fraction minus ln ⁡ open parenthesis 0.2 close parenthesis over 0.4 approximately equals 4.02")[$t = frac(− ln ( 0.2 ), 0.4) ≈ 4.02$] ] ] Solve #math.equation(block: false, alt: "N equals A e to the power minus k t")[$N = A e^(− k t)$] for #math.equation(block: false, alt: "k")[$k$]. #math.equation(block: false, alt: "k equals")[$k =$]\_\_\_\_\_ #math.equation(block: true, alt: "Divide both sides by A .")[$"Divide both sides by" " " bold(italic(A)) .$] #math.equation(block: true, alt: "Take the natural log of both sides.")[$"Take the natural log of both sides."$] #math.equation(block: true, alt: "Divide both sides by minus t .")[$"Divide both sides by" " " − bold(italic(t)) .$] #solutionbox[ #math.equation(block: true, alt: "k equals the fraction minus ln ⁡ open parenthesis N / A close parenthesis over t")[$k = display(frac(− ln ( N / A ), t))$] ] Solve #math.equation(block: false, alt: "N equals A e to the power minus k t")[$N = A e^(− k t)$] for #math.equation(block: false, alt: "k")[$k$]. #solutionbox[ #math.equation(block: true, alt: "k equals the fraction minus ln ⁡ open parenthesis N / A close parenthesis over t")[$k = display(frac(− ln ( N / A ), t))$] ] Delbert says that he will begin solving the equation #math.equation(block: false, alt: "100 e to the power 0.6 t equals 40")[$100 e^(0.6 t) = 40$] by computing #math.equation(block: false, alt: "e to the power 0.6 t")[$e^(0.6 t)$]. Is this a good strategy? Why or why not? \_\_\_\_\_ Delbert says that he will begin solving the equation #math.equation(block: false, alt: "100 e to the power 0.6 t equals 40")[$100 e^(0.6 t) = 40$] by computing #math.equation(block: false, alt: "e to the power 0.6 t")[$e^(0.6 t)$]. Is this a good strategy? Why or why not? === Exponential Growth and Decay In Exponential Growth and Decay, we considered functions of the form #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals P sub 0 times b to the power t")[$P ( t ) = P_(0) ⋅ b^(t)$] which describe exponential growth when #math.equation(block: false, alt: "b greater than 1")[$b > 1$] and exponential decay when #math.equation(block: false, alt: "0 less than b less than 1")[$0 < b < 1$]. Exponential growth and decay can also be modeled by functions of the form #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals P sub 0 times e to the power k t")[$bold(italic(P)) ( bold(italic(t)) ) = bold(italic(P))_(0) ⋅ bold(italic(e))^(bold(italic(k)) bold(italic(t)))$] where we have substituted #math.equation(block: false, alt: "e to the power k")[$e^(k)$] for the growth factor #math.equation(block: false, alt: "b")[$b$], so that #math.equation(block: true, alt: "P open parenthesis t close parenthesis, equals P sub 0 times b to the power t; equals P sub 0 times open parenthesis e to the power k close parenthesis to the power t equals P sub 0 times e to the power k t")[$P ( t ) & = P_(0) ⋅ b^(t) \ & = P_(0) ⋅ attach(( e^(k) ), t: t) = P_(0) ⋅ e^(k t)$] We can find the value of #math.equation(block: false, alt: "k")[$k$] by solving the equation #math.equation(block: false, alt: "b equals e to the power k")[$b = e^(k)$] for #math.equation(block: false, alt: "k")[$k$], to get #math.equation(block: false, alt: "k equals ln ⁡ b")[$k = ln b$]. For instance, consider a colony of bacteria grows according to the formula #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 100 times 3 to the power t")[$P ( t ) = 100 ⋅ 3^(bold(italic(t)))$] We can express this function in the form #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals 100 times e to the power k t")[$P ( t ) = 100 ⋅ bold(italic(e))^(bold(italic(k)) bold(italic(t)))$] if we set #math.equation(block: true, alt: "3 equals e to the power k or k equals ln ⁡ open parenthesis 3 close parenthesis approximately equals 1.0986")[$3 = e^(k) " " " or " " " k = ln ( 3 ) ≈ 1.0986$] Thus, the growth law for the colony of bacteria can be written #math.equation(block: true, alt: "P open parenthesis t close parenthesis approximately equals 100 times e to the power 1.0986 t")[$P ( t ) ≈ 100 ⋅ e^(1.0986 t)$] By graphing both functions on your calculator, you can verify that #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 100 times 3 to the power t; and; P open parenthesis t close parenthesis equals 100 times e to the power 1.0986 t")[$P ( t ) = 100 ⋅ 3^(t) \ " " " " " " "and" " " " " " " \ P ( t ) = 100 ⋅ e^(1.0986 t)$] are just two ways of writing the same function. #figure(figph[P(t) equals 100 3 to the t], alt: "P(t) equals 100 3 to the t", caption: none) Sometimes exponential growth is given as a percentage, so for example we might say “prices rose by 5% annually.” In this case the growth factor is #math.equation(block: false, alt: "b equals 1 plus r")[$" " b = 1 + r " "$], where #math.equation(block: false, alt: "r")[$r$] is the percentage rate in decimal form. For a percent decrease, #math.equation(block: false, alt: "b equals 1 minus r")[$" " b = 1 − r " "$]. #examplebox("Example 8")[][ From 1990 to 2000, the population of Clark County, Nevada, grew by #math.equation(block: false, alt: "6.4 %")[$6.4 upright(%)$] per year. + What was the growth factor for the population of Clark County from 1990 to 2000? If the population of Clark County was #math.equation(block: false, alt: "768 , 000")[$768 , 000$] in 1990, write a formula for the population #math.equation(block: false, alt: "t")[$t$] years later. + Write a growth formula for Clark County using base #math.equation(block: false, alt: "e")[$e$]. #solutionbox[ + The growth factor was #math.equation(block: false, alt: "b equals 1 plus r equals 1.064")[$b = 1 + r = 1.064$]. The population #math.equation(block: false, alt: "t")[$t$] years later was #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 768 , 000 open parenthesis 1.064 close parenthesis to the power t")[$P ( t ) = 768 , 000 ( 1.064 )^(t)$] + We use the formula #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals P sub 0 times e to the power k t")[$P ( t ) = P_(0) ⋅ e^(k t)$], where #math.equation(block: false, alt: "e to the power k equals 1.064")[$e^(k) = 1.064$]. Solving for #math.equation(block: false, alt: "k")[$k$], we find #math.equation(block: true, alt: "k equals ln ⁡ open parenthesis 1.064 close parenthesis equals 0.062")[$k = ln ( 1.064 ) = 0.062$] so #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals 768 , 000 e to the power 0.062 t")[$P ( t ) = 768 , 000 e^(0.062 t)$]. ] ] From 1994 to 1998, the number of personal computers connected to the Internet grew according to the formula #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals 2.8 e to the power 0.85 t")[$N ( t ) = 2.8 e^(0.85 t)$], where #math.equation(block: false, alt: "t equals 0")[$t = 0$] in 1994 and #math.equation(block: false, alt: "N")[$N$] is in millions. (Source: Los Angeles Times, September 6, 1999) + Evaluate #math.equation(block: false, alt: "N open parenthesis 1 close parenthesis equals")[$N ( 1 ) =$]\_\_\_\_\_ . By what percent did the number of Internet users grow in one year? #linebreak() About \_\_\_\_\_% + Express the growth law in the form #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals N sub 0 open parenthesis 1 plus r close parenthesis to the power t")[$N ( t ) = N_(0) ( 1 + r )^(t)$]. #linebreak() #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals")[$N ( t ) =$]\_\_\_\_\_ #math.equation(block: true, alt: "e to the power k equals 1 plus r")[$e^(k) = 1 + r$] #solutionbox[ + #math.equation(block: false, alt: "N open parenthesis 1 close parenthesis approximately equals 6.55")[$N ( 1 ) ≈ 6.55$], #math.equation(block: false, alt: "134 %")[$" " 134 upright(%)$] + #math.equation(block: false, alt: "N open parenthesis t close parenthesis approximately equals 2.8 open parenthesis 2.3396 close parenthesis to the power t")[$N ( t ) ≈ 2.8 ( 2.3396 )^(t)$] ] From 1994 to 1998, the number of personal computers connected to the Internet grew according to the formula #math.equation(block: true, alt: "N open parenthesis t close parenthesis equals 2.8 e to the power 0.85 t")[$N ( t ) = 2.8 e^(0.85 t)$] where #math.equation(block: false, alt: "t equals 0")[$t = 0$] in 1994 and #math.equation(block: false, alt: "N")[$N$] is in millions. (Source: Los Angeles Times, September 6, 1999) + Evaluate #math.equation(block: false, alt: "N open parenthesis 1 close parenthesis")[$N ( 1 )$]. By what percent did the number of Internet users grow in one year? + Express the growth law in the form #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals N sub 0 open parenthesis 1 plus r close parenthesis to the power t")[$" " N ( t ) = N_(0) ( 1 + r )^(t)$]. (Hint: #math.equation(block: false, alt: "e to the power k equals 1 plus r")[$e^(k) = 1 + r$].) #solutionbox[ + #math.equation(block: false, alt: "N open parenthesis 1 close parenthesis approximately equals 6.55")[$N ( 1 ) ≈ 6.55$], #math.equation(block: false, alt: "134 %")[$" " 134 upright(%)$] + #math.equation(block: false, alt: "N open parenthesis t close parenthesis approximately equals 2.8 open parenthesis 2.3396 close parenthesis to the power t")[$N ( t ) ≈ 2.8 ( 2.3396 )^(t)$] ] Now, what about exponential decay, where the the decay factor #math.equation(block: false, alt: "b")[$b$] is a number less than 1? If #math.equation(block: false, alt: "k")[$k$] is negative, then #math.equation(block: false, alt: "e to the power k")[$e^(k)$] is a number less than #math.equation(block: false, alt: "1")[$1$]. For example, if #math.equation(block: false, alt: "k equals minus 2")[$k = − 2$], #math.equation(block: true, alt: "e to the power minus 2 equals the fraction 1 over e squared approximately equals the fraction 1 over 7.3891 approximately equals 0.1353")[$e^(− 2) = frac(1, e^(2)) ≈ frac(1, 7.3891) ≈ 0.1353$] Thus, for negative values of #math.equation(block: false, alt: "k")[$k$], the function #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals P sub 0 e to the power k t")[$P ( t ) = P_(0) e^(k t)$] describes exponential decay. #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 e to the power k t")[$bold(italic(P)) ( bold(italic(t)) ) = bold(italic(P))_(0) bold(italic(e))^(bold(italic(k)) bold(italic(t)))$] describes exponential growth if #math.equation(block: false, alt: "k greater than 0")[$k > 0$], and exponential decay if #math.equation(block: false, alt: "k less than 0")[$k < 0$]. ] The natural log of a fraction between 0 and 1 is \_\_\_\_\_ #solutionbox[ negative. ] The natural log of a number between 0 and 1 is + positive. + negative. + undefined. + between #math.equation(block: false, alt: "e to the power 0")[$e^(0)$] and #math.equation(block: false, alt: "e to the power 1")[$e^(1)$]. #examplebox("Example 9")[][ Express the decay law #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals 60 open parenthesis 0.8 close parenthesis to the power t")[$N ( t ) = 60 ( 0.8 )^(t)$] in the form #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals N sub 0 e to the power k t")[$N ( t ) = N_(0) e^(k t)$]. #solutionbox[ For this decay law, #math.equation(block: false, alt: "N sub 0 equals 60")[$N_(0) = 60$] and #math.equation(block: false, alt: "b equals 0.8")[$b = 0.8$]. We would like to find a value for #math.equation(block: false, alt: "k")[$k$] so that #math.equation(block: false, alt: "e to the power k equals b equals 0.8")[$e^(k) = b = 0.8$], that is, we must solve the equation #math.equation(block: true, alt: "e to the power k, equals 0.8, Take natural log of both sides.; ln ⁡ e to the power k, equals ln ⁡ open parenthesis 0.8 close parenthesis, Simplify.; k, equals ln ⁡ open parenthesis 0.8 close parenthesis approximately equals minus 0.2231")[$e^(k) & = 0.8 & & "Take natural log of both sides." \ ln e^(k) & = ln ( 0.8 ) & & "Simplify." \ k & = ln ( 0.8 ) ≈ − 0.2231$] Replacing #math.equation(block: false, alt: "b")[$b$] with #math.equation(block: false, alt: "e to the power k")[$e^(k)$], we find that the decay law is #math.equation(block: true, alt: "N open parenthesis t close parenthesis approximately equals 60 e to the power minus 0.2231 t")[$N ( t ) ≈ 60 e^(− 0.2231 t)$] ] ] A scientist isolates #math.equation(block: false, alt: "25")[$25$] grams of krypton-91, which decays according to the formula #math.equation(block: true, alt: "N open parenthesis t close parenthesis equals 25 e to the power minus 0.07 t")[$N ( t ) = 25 e^(− 0.07 t)$] , where #math.equation(block: false, alt: "t")[$t$] is in seconds. + Complete the table of values showing the amount of krypton-91 left at #math.equation(block: false, alt: "10")[$10$]-second intervals over the first minute.#figure(table( columns: 5, 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: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "30")[$30$]]), [#math.equation(block: false, alt: "N open parenthesis t close parenthesis")[$N ( t )$]], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], )) #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "60")[$60$]]), [#math.equation(block: false, alt: "N open parenthesis t close parenthesis")[$N ( t )$]], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], )) + Use the table to choose a suitable window and graph the function #math.equation(block: false, alt: "N open parenthesis t close parenthesis")[$N ( t )$]. + Write and solve an equation to answer the question: How long does it take for 60% of the krypton-91 to decay? #linebreak() \_\_\_\_\_ #linebreak() Answer: #math.equation(block: false, alt: "t equals")[$t =$]\_\_\_\_\_ seconds If #math.equation(block: false, alt: "60 %")[$60 upright(%)$] of the krypton-91 has decayed, #math.equation(block: false, alt: "40 %")[$40 upright(%)$] of the original #math.equation(block: false, alt: "25")[$25$] grams remains. #solutionbox[ + #figure(table( columns: 8, 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: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "60")[$60$]]), [#math.equation(block: false, alt: "N open parenthesis t close parenthesis")[$N ( t )$]], [#math.equation(block: false, alt: "25")[$25$]], [#math.equation(block: false, alt: "12.41")[$12.41$]], [#math.equation(block: false, alt: "6.16")[$6.16$]], [#math.equation(block: false, alt: "3.06")[$3.06$]], [#math.equation(block: false, alt: "1.52")[$1.52$]], [#math.equation(block: false, alt: "0.75")[$0.75$]], [#math.equation(block: false, alt: "0.37")[$0.37$]], )) + A graph is below. + #math.equation(block: false, alt: "25 e to the power minus 0.07 t equals 0.40 open parenthesis 25 close parenthesis")[$25 e^(− 0.07 t) = 0.40 ( 25 )$]; #math.equation(block: false, alt: "t equals the fraction ln ⁡ open parenthesis 0.4 close parenthesis over minus 0.07 approximately equals 13.09")[$t = display(frac(ln ( 0.4 ), − 0.07)) ≈ 13.09$] seconds ] Graph for part (b): #figure(figph[decay], alt: "decay", caption: none) A scientist isolates #math.equation(block: false, alt: "25")[$25$] grams of krypton-91, which decays according to the formula #math.equation(block: true, alt: "N open parenthesis t close parenthesis equals 25 e to the power minus 0.07 t")[$N ( t ) = 25 e^(− 0.07 t)$] , where #math.equation(block: false, alt: "t")[$t$] is in seconds. + Complete the table of values showing the amount of krypton-91 left at #math.equation(block: false, alt: "10")[$10$]-second intervals over the first minute.#figure(table( columns: 8, 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: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "60")[$60$]]), [#math.equation(block: false, alt: "N open parenthesis t close parenthesis")[$N ( t )$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + Use the table to choose a suitable window and graph the function #math.equation(block: false, alt: "N open parenthesis t close parenthesis")[$N ( t )$]. + Write and solve an equation to answer the question: How long does it take for 60% of the krypton-91 to decay? #linebreak() Hint: If #math.equation(block: false, alt: "60 %")[$60 upright(%)$] of the krypton-91 has decayed, #math.equation(block: false, alt: "40 %")[$40 upright(%)$] of the original #math.equation(block: false, alt: "25")[$25$] grams remains. #solutionbox[ + #figure(table( columns: 8, 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: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "60")[$60$]]), [#math.equation(block: false, alt: "N open parenthesis t close parenthesis")[$N ( t )$]], [#math.equation(block: false, alt: "25")[$25$]], [#math.equation(block: false, alt: "12.41")[$12.41$]], [#math.equation(block: false, alt: "6.16")[$6.16$]], [#math.equation(block: false, alt: "3.06")[$3.06$]], [#math.equation(block: false, alt: "1.52")[$1.52$]], [#math.equation(block: false, alt: "0.75")[$0.75$]], [#math.equation(block: false, alt: "0.37")[$0.37$]], )) + #figure(figph[exponential decay], alt: "exponential decay", caption: none) + #math.equation(block: false, alt: "25 e to the power minus 0.07 t equals 0.40 open parenthesis 25 close parenthesis")[$25 e^(− 0.07 t) = 0.40 ( 25 )$]; #math.equation(block: false, alt: "t equals the fraction ln ⁡ open parenthesis 0.4 close parenthesis over minus 0.07 approximately equals 13.09")[$" " " " t = display(frac(ln ( 0.4 ), − 0.07)) ≈ 13.09$] seconds ] Explain how to rewrite #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals P sub 0 b to the power t")[$P ( t ) = P_(0) b^(t)$] with the natural base. \_\_\_\_\_ Explain how to rewrite #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals P sub 0 b to the power t")[$P ( t ) = P_(0) b^(t)$] with the natural base. === Continuous Compounding In Section 2.1 we looked at a formula for savings accounts on which the interest is compounded #math.equation(block: false, alt: "n")[$n$] times per year, and we saw that the amount #math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$] on such an account increased when #math.equation(block: false, alt: "n")[$n$] increased. But there is a limit or upper bound to the amount, no matter how large the value of #math.equation(block: false, alt: "n")[$n$]. At this upper bound we say that the interest is #strong[compounded continuously], and the amount is given by the function #math.equation(block: true, alt: "A open parenthesis t close parenthesis equals P e to the power r t")[$A ( t ) = P e^(r t)$] where #math.equation(block: false, alt: "P")[$P$] is the principal invested and #math.equation(block: false, alt: "r")[$r$] is the interest rate. #examplebox("Example 10")[][ Suppose you invest \$#math.equation(block: false, alt: "500")[$500$] in an account that pays #math.equation(block: false, alt: "8 %")[$8 upright(%)$] interest compounded continuously. You leave the money in the account without making any additional deposits or withdrawals. + Write a formula that gives the value of your account #math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$] after #math.equation(block: false, alt: "t")[$t$] years. + Make a table of values showing #math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$] for the first #math.equation(block: false, alt: "5")[$5$] years. + Graph the function #math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$]. + How much will the account be worth after #math.equation(block: false, alt: "10")[$10$] years? + How long will it be before the account is worth \$#math.equation(block: false, alt: "1000")[$1000$]? #solutionbox[ + We substitute #math.equation(block: false, alt: "500")[$500$] for #math.equation(block: false, alt: "P")[$P$], and #math.equation(block: false, alt: "0.08")[$0.08$] for #math.equation(block: false, alt: "r")[$r$] to find #math.equation(block: true, alt: "A open parenthesis t close parenthesis equals 500 e to the power 0.08 t")[$A ( t ) = 500 e^(0.08 t)$] + We evaluate the formula for #math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$] to obtain a table.#figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#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: "500")[$500$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "541.64")[$541.64$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "586.76")[$586.76$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "635.62")[$635.62$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "688.56")[$688.56$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "745.91")[$745.91$]], )) #figure(figph[graph of continuous compounded interest], alt: "graph of continuous compounded interest", caption: none) + The graph of #math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$] is shown above. + We evaluate #math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$] for #math.equation(block: false, alt: "t equals 10")[$t = 10$]. #math.equation(block: true, alt: "A open parenthesis 10 close parenthesis, equals 500 e to the power 0.08 open parenthesis 10 close parenthesis; equals 500 e to the power 0.8; approximately equals 500 open parenthesis 2.2255 close parenthesis equals 1112.77")[$A ( 10 ) & = 500 e^(0.08 ( 10 )) \ & = 500 e^(0.8) \ & ≈ 500 ( 2.2255 ) = 1112.77$] The account will be worth \$#math.equation(block: false, alt: "1112.77")[$1112.77$] after #math.equation(block: false, alt: "10")[$10$] years. + We substitute #math.equation(block: false, alt: "1000")[$1000$] for #math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$] and solve the equation. #math.equation(block: true, alt: "1000, equals 500 e to the power 0.08 t, Divide both sides by 500.; 2, equals e to the power 0.08 t, Take natural log of both sides.; ln ⁡ open parenthesis 2 close parenthesis, equals ln ⁡ open parenthesis e to the power 0.08 t close parenthesis equals 0.08 t, Divide both sides by 0.08.; t, equals the fraction ln ⁡ open parenthesis 2 close parenthesis over 0.08 approximately equals 8.6643")[$1000 & = 500 e^(0.08 t) & & "Divide both sides by 500." \ 2 & = e^(0.08 t) & & "Take natural log of both sides." \ ln ( 2 ) & = ln ( e^(0.08 t) ) = 0.08 t & & "Divide both sides by 0.08." \ t & = frac(ln ( 2 ), 0.08) ≈ 8.6643$] The account will be worth \$#math.equation(block: false, alt: "1000")[$1000$] after approximately #math.equation(block: false, alt: "8.7")[$8.7$] years. ] ] Zelda invested \$1000 in an account that pays 4.5% interest compounded continuously. How long will it be before the account is worth \$2000? Answer: about \_\_\_\_\_ years. #solutionbox[ About 15.4 years ] Zelda invested \$1000 in an account that pays 4.5% interest compounded continuously. How long will it be before the account is worth \$2000? #solutionbox[ About 15.4 years ] Explain why solving exponential equations in base #math.equation(block: false, alt: "e")[$e$] is no harder than solving exponential equations in base 10. \_\_\_\_\_ Explain why solving exponential equations in base #math.equation(block: false, alt: "e")[$e$] is no harder than solving exponential equations in base 10. === Section Summary ==== Vocabulary Look up the definitions of new terms in the Glossary. - Natural exponential function - Natural logarithm - Continuous compounding ==== CONCEPTS + The #strong[natural base] is an irrational number called #math.equation(block: false, alt: "e")[$e$], where #math.equation(block: true, alt: "e approximately equals 2.71828182845")[$e ≈ 2.71828182845$] + The #strong[natural exponential function] is the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals e to the power x")[$f ( x ) = e^(x)$]. The #strong[natural log function] is the function #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals ln ⁡ x equals log sub e ⁡ x")[$g ( x ) = ln x = log_(e) x$]. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Conversion Formulas for Natural Logs] #math.equation(block: true, alt: "y equals ln ⁡ open parenthesis x close parenthesis if and only if e to the power y equals x")[$bold(italic(y)) = bold(ln) ( bold(italic(x)) ) " " " " " if and only if " " " " " bold(italic(e))^(bold(italic(y))) = bold(italic(x))$] ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Properties of Natural Logarithms] If #math.equation(block: false, alt: "x , y greater than 0")[$x , y > 0$], then + #math.equation(block: false, alt: "ln ⁡ open parenthesis x y close parenthesis equals ln ⁡ open parenthesis x close parenthesis plus ln ⁡ open parenthesis y close parenthesis")[$ln ( x y ) = ln ( x ) + ln ( y )$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis the fraction x over y close parenthesis equals ln ⁡ open parenthesis x close parenthesis minus ln ⁡ open parenthesis y close parenthesis")[$ln ( display(frac(x, y)) ) = ln ( x ) − ln ( y )$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis x to the power k close parenthesis equals k ln ⁡ open parenthesis x close parenthesis")[$ln ( x^(k) ) = k ln ( x )$] ] + We use the natural logarithm to solve exponential equations with base #math.equation(block: false, alt: "e")[$e$]. + #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 e to the power k t")[$bold(italic(P)) ( bold(italic(t)) ) = bold(italic(P))_(0) bold(italic(e))^(bold(italic(k)) bold(italic(t)))$]describes exponential growth if #math.equation(block: false, alt: "k greater than 0")[$k > 0$], and exponential decay if #math.equation(block: false, alt: "k less than 0")[$k < 0$]. ] + #strong[Continuous compounding]: The amount accumulated in an account after #math.equation(block: false, alt: "t")[$t$] years at interest rate #math.equation(block: false, alt: "r")[$r$] compounded continuously is given by #math.equation(block: true, alt: "A open parenthesis t close parenthesis equals P e to the power r t")[$A ( t ) = P e^(r t)$] where #math.equation(block: false, alt: "P")[$P$] is the principal invested. ==== STUDY QUESTIONS + State the value of #math.equation(block: false, alt: "e")[$e$] to #math.equation(block: false, alt: "3")[$3$] decimal places. Memorize this value. + Explain why #math.equation(block: false, alt: "ln ⁡ e to the power x equals x")[$ln e^(x) = x$]. + State the formula for exponential growth using base #math.equation(block: false, alt: "e")[$e$]. + How is the formula for exponential decay in base #math.equation(block: false, alt: "e")[$e$] different from the formula for exponential growth? ==== SKILLS Practice each skill in the Homework problems listed. + Graph exponential functions base #math.equation(block: false, alt: "e")[$e$]: \#1–4 + Simplify expressions: \#5 and 6 + Solve exponential and log equations base #math.equation(block: false, alt: "e")[$e$]: \#7–10, 23–30 + Use the properties of logs and exponents with the natural base: \#19–22, 37–40 + Use the natural exponential function in applications: \#11–14, 47–58 + Convert between #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals P sub 0 open parenthesis 1 plus r close parenthesis to the power t")[$P ( t ) = P_(0) ( 1 + r )^(t)$] and #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals P sub 0 e to the power k t")[$P ( t ) = P_(0) e^(k t)$]: \#15–18, 41–46 === Homework 5.3 For Problems 1–4, use your calculator to complete the table for each function. Then choose a suitable window and graph the function. #figure(table( columns: 8, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 10")[$− 10$]], [#math.equation(block: false, alt: "minus 5")[$− 5$]], [#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: "f open parenthesis x close parenthesis")[$f ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals e to the power 0.2 x")[$f ( x ) = e^(0.2 x)$] #figure(table( columns: 8, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 10")[$− 10$]], [#math.equation(block: false, alt: "minus 5")[$− 5$]], [#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: "f open parenthesis x close parenthesis")[$f ( x )$]], [#math.equation(block: false, alt: "0.135")[$0.135$]], [#math.equation(block: false, alt: "0.368")[$0.368$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2.718")[$2.718$]], [#math.equation(block: false, alt: "7.389")[$7.389$]], [#math.equation(block: false, alt: "20.086")[$20.086$]], [#math.equation(block: false, alt: "54.598")[$54.598$]], )) #figure(figph[growth], alt: "growth", caption: none) #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals e to the power 0.6 x")[$f ( x ) = e^(0.6 x)$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals e to the power minus 0.3 x")[$f ( x ) = e^(− 0.3 x)$] #figure(table( columns: 8, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 10")[$− 10$]], [#math.equation(block: false, alt: "minus 5")[$− 5$]], [#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: "f open parenthesis x close parenthesis")[$f ( x )$]], [#math.equation(block: false, alt: "20.086")[$20.086$]], [#math.equation(block: false, alt: "4.482")[$4.482$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0.223")[$0.223$]], [#math.equation(block: false, alt: "0.05")[$0.05$]], [#math.equation(block: false, alt: "0.011")[$0.011$]], [#math.equation(block: false, alt: "0.00248")[$0.00248$]], )) #figure(figph[decay], alt: "decay", caption: none) #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals e to the power minus 0.1 x")[$f ( x ) = e^(− 0.1 x)$] For Problems 5–6, simplify. + #math.equation(block: false, alt: "ln ⁡ open parenthesis e squared close parenthesis")[$ln ( e^(2) )$] + #math.equation(block: false, alt: "e to the power ln ⁡ open parenthesis 5 t close parenthesis")[$e^(ln ( 5 t ))$] + #math.equation(block: false, alt: "e to the power minus ln ⁡ open parenthesis x close parenthesis")[$e^(− ln ( x ))$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis the square root of e close parenthesis")[$ln ( sqrt(e) )$] + #math.equation(block: false, alt: "2")[$2$] + #math.equation(block: false, alt: "5 t")[$5 t$] + #math.equation(block: false, alt: "the fraction 1 over x")[$display(frac(1, x))$] + #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis e to the power x to the power 4 close parenthesis")[$ln ( e^(x^(4)) )$] + #math.equation(block: false, alt: "e to the power 3 ln ⁡ open parenthesis x close parenthesis")[$e^(3 ln ( x ))$] + #math.equation(block: false, alt: "e to the power ln ⁡ open parenthesis x close parenthesis minus ln ⁡ open parenthesis y close parenthesis")[$e^(ln ( x ) − ln ( y ))$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis the fraction 1 over e to the power 2 t close parenthesis")[$ln ( display(frac(1, e^(2 t))) )$] For Problems 7–10, solve for #math.equation(block: false, alt: "x")[$x$]. Round your answers to two decimal places. + #math.equation(block: false, alt: "e to the power x equals 1.9")[$e^(x) = 1.9$] + #math.equation(block: false, alt: "e to the power x equals 45")[$e^(x) = 45$] + #math.equation(block: false, alt: "e to the power x equals 0.3")[$e^(x) = 0.3$] + #math.equation(block: false, alt: "0.64")[$0.64$] + #math.equation(block: false, alt: "3.81")[$3.81$] + #math.equation(block: false, alt: "minus 1.20")[$− 1.20$] + #math.equation(block: false, alt: "e to the power x equals 2.1")[$e^(x) = 2.1$] + #math.equation(block: false, alt: "e to the power x equals 60")[$e^(x) = 60$] + #math.equation(block: false, alt: "e to the power x equals 0.9")[$e^(x) = 0.9$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis x close parenthesis equals 1.42")[$ln ( x ) = 1.42$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis x close parenthesis equals 0.63")[$ln ( x ) = 0.63$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis x close parenthesis equals minus 2.6")[$ln ( x ) = − 2.6$] + #math.equation(block: false, alt: "4.14")[$4.14$] + #math.equation(block: false, alt: "1.88")[$1.88$] + #math.equation(block: false, alt: "0.07")[$0.07$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis x close parenthesis equals 2.03")[$ln ( x ) = 2.03$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis x close parenthesis equals 0.59")[$ln ( x ) = 0.59$] + #math.equation(block: false, alt: "ln ⁡ open parenthesis x close parenthesis equals minus 3.4")[$ln ( x ) = − 3.4$] The number of bacteria in a culture grows according to the function #math.equation(block: true, alt: "N open parenthesis t close parenthesis equals N sub 0 e to the power 0.04 t")[$N ( t ) = N_(0) e^(0.04 t)$] where #math.equation(block: false, alt: "N sub 0")[$N_(0)$] is the number of bacteria present at time #math.equation(block: false, alt: "t equals 0")[$t = 0$] and #math.equation(block: false, alt: "t")[$t$] is the time in hours. + Write a growth law for a sample in which #math.equation(block: false, alt: "6000")[$6000$] bacteria were present initially. + Make a table of values for #math.equation(block: false, alt: "N open parenthesis t close parenthesis")[$N ( t )$] in #math.equation(block: false, alt: "5")[$5$]-hour intervals over the first #math.equation(block: false, alt: "30")[$30$] hours. + Graph #math.equation(block: false, alt: "N open parenthesis t close parenthesis")[$N ( t )$]. + How many bacteria were present at #math.equation(block: false, alt: "t equals 24")[$t = 24$] hours? + How much time must elapse (to the nearest tenth of an hour) for the original #math.equation(block: false, alt: "6000")[$6000$] bacteria to increase to #math.equation(block: false, alt: "100 , 000")[$100 , 000$]? + #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals 6000 e to the power 0.04 t")[$N ( t ) = 6000 e^(0.04 t)$] + #figure(table( columns: 8, 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: "30")[$30$]]), [#math.equation(block: false, alt: "N open parenthesis t close parenthesis")[$N ( t )$]], [#math.equation(block: false, alt: "6000")[$6000$]], [#math.equation(block: false, alt: "7328")[$7328$]], [#math.equation(block: false, alt: "8951")[$8951$]], [#math.equation(block: false, alt: "10 , 933")[$10 , 933$]], [#math.equation(block: false, alt: "13 , 353")[$13 , 353$]], [#math.equation(block: false, alt: "16 , 310")[$16 , 310$]], [#math.equation(block: false, alt: "19 , 921")[$19 , 921$]], )) + #figure(figph[growth], alt: "growth", caption: none) + #math.equation(block: false, alt: "15 , 670")[$15 , 670$] + #math.equation(block: false, alt: "70.3")[$70.3$] hrs Hope invests #math.equation(block: false, alt: "$ 2000")[$upright(\$) 2000$] in a savings account that pays #math.equation(block: false, alt: "5 the fraction 1 over 2 %")[$5 frac(1, 2) upright(%)$] annual interest compounded continuously. + Write a formula that gives the amount of money #math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$] in Hope’s account after #math.equation(block: false, alt: "t")[$t$] years. + Make a table of values for #math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$] in #math.equation(block: false, alt: "2")[$2$]-year intervals over the first #math.equation(block: false, alt: "10")[$10$] years. + Graph #math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$]. + How much will Hope's account be worth after #math.equation(block: false, alt: "7")[$7$] years? + How long will it take for the account to grow to #math.equation(block: false, alt: "$ 5000")[$upright(\$) 5000$]? The intensity, #math.equation(block: false, alt: "I")[$I$] (in lumens), of a light beam after passing through #math.equation(block: false, alt: "t")[$t$] centimeters of a filter having an absorption coefficient of #math.equation(block: false, alt: "0.1")[$0.1$] is given by the function #math.equation(block: true, alt: "I open parenthesis t close parenthesis equals 1000 e to the power minus 0.1 t")[$I ( t ) = 1000 e^(− 0.1 t)$] + Graph #math.equation(block: false, alt: "I open parenthesis t close parenthesis")[$I ( t )$]. + What is the intensity (to the nearest tenth of a lumen) of a light beam that has passed through #math.equation(block: false, alt: "0.6")[$0.6$] centimeter of the filter? + How many centimeters (to the nearest tenth) of the filter will reduce the illumination to #math.equation(block: false, alt: "800")[$800$] lumens? + #figure(figph[decay], alt: "decay", caption: none) + #math.equation(block: false, alt: "941.8")[$941.8$] lumens + #math.equation(block: false, alt: "2.2")[$2.2$] cm X-rays can be absorbed by a lead plate so that #math.equation(block: true, alt: "I open parenthesis t close parenthesis equals I sub 0 e to the power minus 1.88 t")[$I ( t ) = I_(0) e^(− 1.88 t)$] where #math.equation(block: false, alt: "I sub 0")[$I_(0)$] is the X-ray count at the source and #math.equation(block: false, alt: "I open parenthesis t close parenthesis")[$I ( t )$] is the X-ray count behind a lead plate of thickness #math.equation(block: false, alt: "t")[$t$] inches. + Graph #math.equation(block: false, alt: "I open parenthesis t close parenthesis")[$I ( t )$]. + What percent of an X-ray beam will penetrate a lead plate #math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$] inch thick? + How thick should the lead plate be in order to screen out #math.equation(block: false, alt: "70 %")[$70 upright(%)$] of the X-rays? For problems 15–18, express each exponential function in the form #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 the function increasing or decreasing? What is its initial value? #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 20 e to the power 0.4 t")[$P ( t ) = 20 e^(0.4 t)$] #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals 20 open parenthesis e to the power 0.4 close parenthesis to the power t approximately equals 20 times 1.492 to the power t")[$P ( t ) = 20 attach(( e^(0.4) ), t: t) ≈ 20 ⋅ 1.492^(t)$]; increasing; initial value #math.equation(block: false, alt: "20")[$20$] #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 0.8 e to the power 1.3 t")[$P ( t ) = 0.8 e^(1.3 t)$] #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 6500 e to the power minus 2.5 t")[$P ( t ) = 6500 e^(− 2.5 t)$] #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals 6500 open parenthesis e to the power minus 2.5 close parenthesis to the power t approximately equals 6500 times 0.082 to the power t")[$P ( t ) = 6500 attach(( e^(− 2.5) ), t: t) ≈ 6500 ⋅ 0.082^(t)$]; decreasing; initial value #math.equation(block: false, alt: "6500")[$6500$] #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 1.7 e to the power minus 0.02 t")[$P ( t ) = 1.7 e^(− 0.02 t)$] + Fill in the table, rounding your answers to four decimal places.#figure(table( columns: 7, 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: "0.5")[$0.5$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "1.5")[$1.5$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2.5")[$2.5$]]), [#math.equation(block: false, alt: "e to the power x")[$e^(x)$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + Compute the ratio of each function value to the previous one. Explain the result. + #figure(table( columns: 7, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "1.5")[$1.5$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2.5")[$2.5$]]), [#math.equation(block: false, alt: "e to the power x")[$e^(x)$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "1.6487")[$1.6487$]], [#math.equation(block: false, alt: "2.7183")[$2.7183$]], [#math.equation(block: false, alt: "4.4817")[$4.4817$]], [#math.equation(block: false, alt: "7.3891")[$7.3891$]], [#math.equation(block: false, alt: "12.1825")[$12.1825$]], )) + Each ratio is #math.equation(block: false, alt: "e to the power 0.5 approximately equals 1.6487")[$e^(0.5) ≈ 1.6487$]: Increasing #math.equation(block: false, alt: "x")[$x$]-values by a constant #math.equation(block: false, alt: "Δ x equals 0.5")[$upright(Δ) x = 0.5$] corresponds to multiplying the #math.equation(block: false, alt: "y")[$y$]-values of the exponential function by a constant factor of #math.equation(block: false, alt: "e to the power Δ x")[$e^(upright(Δ) x)$]. + Fill in the table, rounding your answers to four decimal places.#figure(table( columns: 7, 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: "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: "e to the power x")[$e^(x)$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + Compute the ratio of each function value to the previous one. Explain the result. + Fill in the table, rounding your answers to the nearest integer.#figure(table( columns: 8, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0.6931")[$0.6931$]], [#math.equation(block: false, alt: "1.3863")[$1.3863$]], [#math.equation(block: false, alt: "2.0794")[$2.0794$]], [#math.equation(block: false, alt: "2.7726")[$2.7726$]], [#math.equation(block: false, alt: "3.4657")[$3.4657$]], [#math.equation(block: false, alt: "4.1589")[$4.1589$]]), [#math.equation(block: false, alt: "e to the power x")[$e^(x)$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + Subtract each #math.equation(block: false, alt: "x")[$x$]-value from the next one. Explain the result. + #figure(table( columns: 8, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0.6931")[$0.6931$]], [#math.equation(block: false, alt: "1.3863")[$1.3863$]], [#math.equation(block: false, alt: "2.0794")[$2.0794$]], [#math.equation(block: false, alt: "2.7726")[$2.7726$]], [#math.equation(block: false, alt: "3.4657")[$3.4657$]], [#math.equation(block: false, alt: "4.1589")[$4.1589$]]), [#math.equation(block: false, alt: "e to the power x")[$e^(x)$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "16")[$16$]], [#math.equation(block: false, alt: "32")[$32$]], [#math.equation(block: false, alt: "64")[$64$]], )) + Each difference in #math.equation(block: false, alt: "x")[$x$]-values is approximately #math.equation(block: false, alt: "ln ⁡ open parenthesis 2 close parenthesis approximately equals 0.6931")[$ln ( 2 ) ≈ 0.6931$]: Increasing #math.equation(block: false, alt: "x")[$x$]-values by a constant #math.equation(block: false, alt: "Δ x equals ln ⁡ open parenthesis 2 close parenthesis")[$upright(Δ) x = ln ( 2 )$] corresponds to multiplying the #math.equation(block: false, alt: "y")[$y$]-values of the exponential function by a constant factor of #math.equation(block: false, alt: "e to the power Δ x equals e to the power ln ⁡ open parenthesis 2 close parenthesis equals 2")[$e^(upright(Δ) x) = e^(ln ( 2 )) = 2$]. That is, each function value is approximately equal to double the previous one. + Fill in the table, rounding your answers to the nearest integer.#figure(table( columns: 8, 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: "1.0986")[$1.0986$]], [#math.equation(block: false, alt: "2.1972")[$2.1972$]], [#math.equation(block: false, alt: "3.2958")[$3.2958$]], [#math.equation(block: false, alt: "4.3944")[$4.3944$]], [#math.equation(block: false, alt: "5.4931")[$5.4931$]], [#math.equation(block: false, alt: "6.5917")[$6.5917$]]), [#math.equation(block: false, alt: "e to the power x")[$e^(x)$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + Subtract each #math.equation(block: false, alt: "x")[$x$]-value from the next one. Explain the result. For Problems 23–30, solve. Round your answers to two decimal places. #math.equation(block: true, alt: "6.21 equals 2.3 e to the power 1.2 x")[$6.21 = 2.3 e^(1.2 x)$] #math.equation(block: true, alt: "0.8277")[$0.8277$] #math.equation(block: true, alt: "22.26 equals 5.3 e to the power 0.4 x")[$22.26 = 5.3 e^(0.4 x)$] #math.equation(block: true, alt: "6.4 equals 20 e to the power 0.3 x minus 1.8")[$6.4 = 20 e^(0.3 x) − 1.8$] #math.equation(block: true, alt: "minus 2.9720")[$− 2.9720$] #math.equation(block: true, alt: "4.5 equals 4 e to the power 2.1 x plus 3.3")[$4.5 = 4 e^(2.1 x) + 3.3$] #math.equation(block: true, alt: "46.52 equals 3.1 e to the power 1.2 x plus 24.2")[$46.52 = 3.1 e^(1.2 x) + 24.2$] #math.equation(block: true, alt: "1.6451")[$1.6451$] #math.equation(block: true, alt: "1.23 equals 1.3 e to the power 2.1 x minus 17.1")[$1.23 = 1.3 e^(2.1 x) − 17.1$] #math.equation(block: true, alt: "16.24 equals 0.7 e to the power minus 1.3 x minus 21.7")[$16.24 = 0.7 e^(− 1.3 x) − 21.7$] #math.equation(block: true, alt: "minus 3.0713")[$− 3.0713$] #math.equation(block: true, alt: "55.68 equals 0.6 e to the power minus 0.7 x plus 23.1")[$55.68 = 0.6 e^(− 0.7 x) + 23.1$] For Problems 31–36, solve the equation for the specified variable. #math.equation(block: false, alt: "y equals e to the power k t ,")[$y = e^(k t) , " " " "$] for #math.equation(block: false, alt: "t")[$t$] #math.equation(block: true, alt: "t equals the fraction 1 over k ln ⁡ open parenthesis y close parenthesis")[$t = display(frac(1, k)) ln ( y )$] #math.equation(block: false, alt: "the fraction T over R equals e to the power t / 2 ,")[$display(frac(T, R)) = e^(t / 2) , " " " "$] for #math.equation(block: false, alt: "t")[$t$] #math.equation(block: false, alt: "y equals k open parenthesis 1 minus e to the power minus t close parenthesis ,")[$y = k ( 1 − e^(− t) ) , " " " "$] for #math.equation(block: false, alt: "t")[$t$] #math.equation(block: true, alt: "t equals ln ⁡ open parenthesis the fraction k over k minus y close parenthesis")[$t = ln ( display(frac(k, k − y)) )$] #math.equation(block: false, alt: "B minus 2 equals open parenthesis A plus 3 close parenthesis e to the power minus t / 3 ,")[$B − 2 = ( A + 3 ) e^(− t / 3) , " " " "$] for #math.equation(block: false, alt: "t")[$t$] #math.equation(block: false, alt: "T equals T sub 0 ln ⁡ open parenthesis k plus 10 close parenthesis ,")[$T = T_(0) ln ( k + 10 ) , " " " "$] for #math.equation(block: false, alt: "k")[$k$] #math.equation(block: true, alt: "k equals e to the power T / T sub 0 minus 10")[$k = e^(T / T_(0)) − 10$] #math.equation(block: false, alt: "P equals P sub 0 plus ln ⁡ open parenthesis 10 k close parenthesis ,")[$P = P_(0) + ln ( 10 k ) , " " " "$] for #math.equation(block: false, alt: "k")[$k$] + Fill in the table, rounding your answers to three decimal places.#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "0.39")[$0.39$]], [#math.equation(block: false, alt: "3.9")[$3.9$]], [#math.equation(block: false, alt: "39")[$39$]], [#math.equation(block: false, alt: "390")[$390$]]), [#math.equation(block: false, alt: "ln ⁡ open parenthesis n close parenthesis")[$ln ( n )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Subtract each natural logarithm in your table from the next one. (For example, compute #math.equation(block: false, alt: "ln ⁡ open parenthesis 3.9 close parenthesis minus ln ⁡ open parenthesis 0.39 close parenthesis")[$ln ( 3.9 ) − ln ( 0.39 )$].) Explain the result. + #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "0.39")[$0.39$]], [#math.equation(block: false, alt: "3.9")[$3.9$]], [#math.equation(block: false, alt: "39")[$39$]], [#math.equation(block: false, alt: "390")[$390$]]), [#math.equation(block: false, alt: "ln ⁡ n")[$ln n$]], [#math.equation(block: false, alt: "minus 0.942")[$− 0.942$]], [#math.equation(block: false, alt: "1.361")[$1.361$]], [#math.equation(block: false, alt: "3.664")[$3.664$]], [#math.equation(block: false, alt: "5.966")[$5.966$]], )) + Each difference in function values is approximately #math.equation(block: false, alt: "ln ⁡ open parenthesis 10 close parenthesis approximately equals 2.303")[$ln ( 10 ) ≈ 2.303$]: Multiplying #math.equation(block: false, alt: "x")[$x$]-values by a constant factor of #math.equation(block: false, alt: "10")[$10$] corresponds to adding a constant value of ln (10) to the #math.equation(block: false, alt: "y")[$y$]-values of the natural log function. + Fill in the table, rounding your answers to three decimal places.#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "0.64")[$0.64$]], [#math.equation(block: false, alt: "6.4")[$6.4$]], [#math.equation(block: false, alt: "64")[$64$]], [#math.equation(block: false, alt: "640")[$640$]]), [#math.equation(block: false, alt: "ln ⁡ open parenthesis n close parenthesis")[$ln ( n )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Subtract each natural logarithm in your table from the next one. (For example, compute #math.equation(block: false, alt: "ln ⁡ open parenthesis 6.4 close parenthesis minus ln ⁡ open parenthesis 0.64 close parenthesis")[$ln ( 6.4 ) − ln ( 0.64 )$].) Explain the result. + Fill in the table, rounding your answers to three decimal places.#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "16")[$16$]]), [#math.equation(block: false, alt: "ln ⁡ open parenthesis n close parenthesis")[$ln ( n )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Divide each natural logarithm in your table by #math.equation(block: false, alt: "ln ⁡ 2")[$ln 2$]. Explain the result. + #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "16")[$16$]]), [#math.equation(block: false, alt: "ln ⁡ open parenthesis n close parenthesis")[$ln ( n )$]], [#math.equation(block: false, alt: "0.693")[$0.693$]], [#math.equation(block: false, alt: "1.386")[$1.386$]], [#math.equation(block: false, alt: "2.079")[$2.079$]], [#math.equation(block: false, alt: "2.773")[$2.773$]], )) + Each quotient equals #math.equation(block: false, alt: "k")[$k$], where #math.equation(block: false, alt: "n equals 2 to the power k")[$n = 2^(k)$]. Because #math.equation(block: false, alt: "ln ⁡ open parenthesis n close parenthesis equals ln ⁡ open parenthesis 2 to the power k close parenthesis equals k times ln ⁡ open parenthesis 2 close parenthesis")[$ln ( n ) = ln ( 2^(k) ) = k ⋅ ln ( 2 )$], #math.equation(block: false, alt: "k equals the fraction ln ⁡ open parenthesis n close parenthesis over ln ⁡ open parenthesis 2 close parenthesis")[$k = display(frac(ln ( n ), ln ( 2 )))$]. + Fill in the table, rounding your answers to three decimal places.#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "25")[$25$]], [#math.equation(block: false, alt: "125")[$125$]], [#math.equation(block: false, alt: "625")[$625$]]), [#math.equation(block: false, alt: "ln ⁡ open parenthesis n close parenthesis")[$ln ( n )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Divide each natural logarithm in your table by #math.equation(block: false, alt: "ln ⁡ open parenthesis 5 close parenthesis")[$ln ( 5 )$]. Explain the result. For Problems 41–46, + Express each growth or decay law in the form #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals N sub 0 e to the power k t")[$N ( t ) = N_(0) e^(k t)$]. + Check your answer by graphing both forms of the function on the same axes. Do they have the same graph? #math.equation(block: true, alt: "N open parenthesis t close parenthesis equals 100 times 2 to the power t")[$N ( t ) = 100 ⋅ 2^(t)$] + #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals 100 e to the power open parenthesis ln ⁡ open parenthesis 2 close parenthesis close parenthesis t approximately equals 100 e to the power 0.6931 t")[$N ( t ) = 100 e^(( ln ( 2 ) ) t) ≈ 100 e^(0.6931 t)$] + #figure(figph[growth], alt: "growth", caption: none) #math.equation(block: true, alt: "N open parenthesis t close parenthesis equals 50 times 3 to the power t")[$N ( t ) = 50 ⋅ 3^(t)$] #math.equation(block: true, alt: "N open parenthesis t close parenthesis equals 1200 open parenthesis 0.6 close parenthesis to the power t")[$N ( t ) = 1200 ( 0.6 )^(t)$] + #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals 1200 e to the power ln ⁡ open parenthesis 0.6 close parenthesis t approximately equals 1200 e to the power minus 0.5108 t")[$N ( t ) = 1200 e^(ln ( 0.6 ) t) ≈ 1200 e^(− 0.5108 t)$] + #figure(figph[decay], alt: "decay", caption: none) #math.equation(block: true, alt: "N open parenthesis t close parenthesis equals 300 open parenthesis 0.8 close parenthesis to the power t")[$N ( t ) = 300 ( 0.8 )^(t)$] #math.equation(block: true, alt: "N open parenthesis t close parenthesis equals 10 open parenthesis 1.15 close parenthesis to the power t")[$N ( t ) = 10 ( 1.15 )^(t)$] + #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals 10 e to the power ln ⁡ open parenthesis 1.15 close parenthesis t approximately equals 10 e to the power 0.1398 t")[$N ( t ) = 10 e^(ln ( 1.15 ) t) ≈ 10 e^(0.1398 t)$] + #figure(figph[growth], alt: "growth", caption: none) #math.equation(block: true, alt: "N open parenthesis t close parenthesis equals 1000 open parenthesis 1.04 close parenthesis to the power t")[$N ( t ) = 1000 ( 1.04 )^(t)$] The population of Citrus Valley was #math.equation(block: false, alt: "20 , 000")[$20 , 000$] in #math.equation(block: false, alt: "2000")[$2000$]. In #math.equation(block: false, alt: "2010")[$2010$], it was #math.equation(block: false, alt: "35 , 000")[$35 , 000$]. + What is #math.equation(block: false, alt: "P sub 0")[$P_(0)$] if #math.equation(block: false, alt: "t equals 0")[$t = 0$] in #math.equation(block: false, alt: "2000")[$2000$]? + Use the population in #math.equation(block: false, alt: "2010")[$2010$] to find the growth factor #math.equation(block: false, alt: "e to the power k")[$e^(k)$]. + Write a growth law of the form #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals P sub 0 e to the power k t")[$P ( t ) = P_(0) e^(k t)$] for the population of Citrus Valley. + If it continues at the same rate of growth, what will the population be in #math.equation(block: false, alt: "2030")[$2030$]? + #math.equation(block: false, alt: "20 , 000")[$20 , 000$] + #math.equation(block: false, alt: "open parenthesis the fraction 35 , 000 over 20 , 000 close parenthesis to the power 1 / 10 approximately equals e to the power 0.056")[$attach(( display(frac(35 "," 000, 20 "," 000)) ), t: 1 / 10) ≈ e^(0.056)$] + #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals 20 , 000 e to the power 0.056 t")[$P ( t ) = 20 , 000 e^(0.056 t)$] + #math.equation(block: false, alt: "107 , 188")[$107 , 188$] A copy of #emph[Time] magazine cost \$#math.equation(block: false, alt: "1.50")[$1.50$] in #math.equation(block: false, alt: "1981.")[$1981. " "$] In #math.equation(block: false, alt: "1988")[$1988$], the cover price had increased to \$#math.equation(block: false, alt: "2.00")[$2.00$]. + What is #math.equation(block: false, alt: "P sub 0")[$P_(0)$] if #math.equation(block: false, alt: "t equals 0")[$t = 0$] in #math.equation(block: false, alt: "1981")[$1981$]? + Use the price in #math.equation(block: false, alt: "1988")[$1988$] to find the growth factor #math.equation(block: false, alt: "e to the power k")[$e^(k)$]. + Find a growth law of the form #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals P sub 0 e to the power k t")[$P ( t ) = P_(0) e^(k t)$] for the price of #emph[Time]. + In #math.equation(block: false, alt: "1999")[$1999$], a copy of #emph[Time] cost \$#math.equation(block: false, alt: "3.50")[$3.50$]. Did the price of the magazine continue to grow at the same rate from #math.equation(block: false, alt: "1981")[$1981$] to #math.equation(block: false, alt: "1999")[$1999$]? Cobalt-60 is a radioactive isotope used in the treatment of cancer. A #math.equation(block: false, alt: "500")[$500$]-milligram sample of cobalt-60 decays to #math.equation(block: false, alt: "385")[$385$] milligrams after #math.equation(block: false, alt: "2")[$2$] years. + Using #math.equation(block: false, alt: "P sub 0 equals 500")[$P_(0) = 500$], find the decay factor #math.equation(block: false, alt: "e to the power k")[$e^(k)$] for cobalt-60. + Write a decay law #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals N sub 0 e to the power k t")[$N ( t ) = N_(0) e^(k t)$] for cobalt-60. + How much of the original sample will be left after #math.equation(block: false, alt: "10")[$10$] years? + #math.equation(block: false, alt: "open parenthesis the fraction 385 over 500 close parenthesis to the power 1 / 2 approximately equals e to the power minus 0.1307")[$attach(( display(frac(385, 500)) ), t: 1 / 2) ≈ e^(− 0.1307)$] + #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals 500 e to the power minus 0.1307 t")[$N ( t ) = 500 e^(− 0.1307 t)$] + #math.equation(block: false, alt: "135.3")[$135.3$] mg Weed seeds can survive for a number of years in the soil. An experiment on cultivated land found #math.equation(block: false, alt: "155")[$155$] million weed seeds per acre, and in the following years the experimenters prevented the seeds from coming to maturity and producing new weeds. Four years later, there were #math.equation(block: false, alt: "13.6")[$13.6$] million seeds per acre. (Source: Burton, 1998) + Find the annual decay factor #math.equation(block: false, alt: "e to the power k")[$e^(k)$] for the number of weed seeds in the soil. + Write an exponential formula with base #math.equation(block: false, alt: "e")[$e$] for the number of weed seeds that survived after #math.equation(block: false, alt: "t")[$t$] years. Problems 51–58 are about doubling time and half-life. Delbert invests \$#math.equation(block: false, alt: "500")[$500$] in an account that pays #math.equation(block: false, alt: "9.5 %")[$9.5 upright(%)$] interest compounded continuously. + Write a formula for #math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$] that gives the amount of money in Delbert's account after #math.equation(block: false, alt: "t")[$t$] years. + How long will it take Delbert's investment to double to \$#math.equation(block: false, alt: "1000")[$1000$]? + How long will it take Delbert's money to double again, to \$#math.equation(block: false, alt: "2000")[$2000$]? + Graph #math.equation(block: false, alt: "A open parenthesis t close parenthesis")[$A ( t )$] and illustrate the doubling time on your graph. + Choose any point #math.equation(block: false, alt: "open parenthesis t sub 1 , A sub 1 close parenthesis")[$( t_(1) , A_(1) )$] on the graph, then find the point on the graph with vertical coordinate #math.equation(block: false, alt: "2 A sub 1")[$2 A_(1)$]. Verify that the difference in the #math.equation(block: false, alt: "t")[$t$]-coordinates of the two points is the doubling time. + #math.equation(block: false, alt: "A open parenthesis t close parenthesis equals 500 e to the power 0.095 t")[$A ( t ) = 500 e^(0.095 t)$] + #math.equation(block: false, alt: "7.3")[$7.3$] years + #math.equation(block: false, alt: "7.3")[$7.3$] years d–e #figure(figph[growth with marked doubling time], alt: "growth with marked doubling time", caption: none) The growth of plant populations can be measured by the amount of pollen they produce. The pollen from a population of pine trees that lived more than #math.equation(block: false, alt: "9500")[$9500$] years ago in Norfolk, England, was deposited in the layers of sediment in a lake basin and dated with radiocarbon techniques. #figure(figph[growth], alt: "growth", caption: none) The figure shows the rate of pollen accumulation plotted against time, and the fitted curve #math.equation(block: false, alt: "P open parenthesis t close parenthesis equals 650 e to the power 0.00932 t")[$P ( t ) = 650 e^(0.00932 t)$]. (Source: Burton, 1998) + What was the annual rate of growth in pollen accumulation? + Find the doubling time for the pollen accumulation, that is, the time it took for the accumulation rate to double. + By what factor did the pollen accumulation rate increase over a period of #math.equation(block: false, alt: "500")[$500$] years? Technetium-99m (Tc-99m) is an artificially produced radionuclide used as a tracer for producing images of internal organs such as the heart, liver, and thyroid. A solution of Tc-99m with initial radioactivity of #math.equation(block: false, alt: "10 , 000")[$10 , 000$] becquerels (Bq) decays according to the formula #math.equation(block: true, alt: "N open parenthesis t close parenthesis equals 10 , 000 e to the power minus 0.1155 t")[$N ( t ) = 10 , 000 e^(− 0.1155 t)$] where #math.equation(block: false, alt: "t")[$t$] is in hours. + How long will it take the radioactivity to fall to half its initial value, or #math.equation(block: false, alt: "5000")[$5000$] Bq? + How long will it take the radioactivity to be halved again? + Graph #math.equation(block: false, alt: "N open parenthesis t close parenthesis")[$N ( t )$] and illustrate the half-life on your graph. + Choose any point #math.equation(block: false, alt: "open parenthesis t sub 1 , N sub 1 close parenthesis")[$( t_(1) , N_(1) )$] on the graph, then find the point on the graph with vertical coordinate #math.equation(block: false, alt: "0.5 N sub 1")[$0.5 N_(1)$]. Verify that the difference in the #math.equation(block: false, alt: "t")[$t$]-coordinates of the two points is the half-life. + #math.equation(block: false, alt: "6")[$6$] hours + #math.equation(block: false, alt: "6")[$6$] hours + #figure(figph[decay with marked half-life], alt: "decay with marked half-life", caption: none) All living things contain a certain amount of the isotope carbon-14. When an organism dies, the carbon-14 decays according to the formula #math.equation(block: true, alt: "N open parenthesis t close parenthesis equals N sub 0 e to the power minus 0.000124 t")[$N ( t ) = N_(0) e^(− 0.000124 t)$] where #math.equation(block: false, alt: "t")[$t$] is measured in years. Scientists can estimate the age of an organic object by measuring the amount of carbon-14 remaining. + When the Dead Sea scrolls were discovered in 1947, they had #math.equation(block: false, alt: "78.8 %")[$78.8 upright(%)$] of their original carbon-14. How old were the Dead Sea scrolls then? + What is the half-life of carbon-14, that is, how long does it take for half of an object's carbon-14 to decay? The half-life of iodine-131 is approximately #math.equation(block: false, alt: "8")[$8$] days. + If a sample initially contains #math.equation(block: false, alt: "N sub 0")[$N_(0)$] grams of iodine-131, how much will it contain after #math.equation(block: false, alt: "8")[$8$] days? How much will it contain after #math.equation(block: false, alt: "16")[$16$] days? After #math.equation(block: false, alt: "32")[$32$] days? + Use your answers to part (a) to sketch a graph of #math.equation(block: false, alt: "N open parenthesis t close parenthesis")[$N ( t )$], the amount of iodine-131 remaining, versus time. (Choose an arbitrary height for #math.equation(block: false, alt: "N sub 0")[$N_(0)$] on the vertical axis.) + Calculate #math.equation(block: false, alt: "k")[$k$], and hence find a decay law of the form #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals N sub 0 e to the power k t")[$N ( t ) = N_(0) e^(k t)$], where #math.equation(block: false, alt: "k less than 0")[$k < 0$], for iodine-131. + #math.equation(block: false, alt: "the fraction 1 over 2 N sub 0")[$frac(1, 2) N_(0)$], #math.equation(block: false, alt: "the fraction 1 over 4 N sub 0")[$frac(1, 4) N_(0)$], #math.equation(block: false, alt: "the fraction 1 over 16 N sub 0")[$frac(1, 16) N_(0)$] + #figure(figph[decay], alt: "decay", caption: none) + #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals N sub 0 e to the power minus 0.0866 t")[$N ( t ) = N_(0) e^(− 0.0866 t)$] The half-life of hydrogen-3 is #math.equation(block: false, alt: "12.5")[$12.5$] years. + If a sample initially contains #math.equation(block: false, alt: "N sub 0")[$N_(0)$] grams of hydrogen-3, how much will it contain after #math.equation(block: false, alt: "12.5")[$12.5$] years? How much will it contain after #math.equation(block: false, alt: "25")[$25$] years? + Use your answers to part (a) to sketch a graph of #math.equation(block: false, alt: "N open parenthesis t close parenthesis")[$N ( t )$], the amount of hydrogen-3 remaining, versus time. (Choose an arbitrary height for #math.equation(block: false, alt: "N sub 0")[$N_(0)$] on the vertical axis.) + Calculate #math.equation(block: false, alt: "k")[$k$], and hence find a decay law of the form #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals N sub 0 e to the power k t")[$N ( t ) = N_(0) e^(k t)$], where #math.equation(block: false, alt: "k less than 0")[$k < 0$], for hydrogen-3. A Geiger counter measures the amount of radioactive material present in a substance. The table shows the count rate for a sample of iodine-128 as a function of time. (Source: Hunt and Sykes, 1984) #figure(table( columns: 11, align: left, inset: 6pt, table.header([Time (min)], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "60")[$60$]], [#math.equation(block: false, alt: "70")[$70$]], [#math.equation(block: false, alt: "80")[$80$]], [#math.equation(block: false, alt: "90")[$90$]]), [Counts/sec], [#math.equation(block: false, alt: "120")[$120$]], [#math.equation(block: false, alt: "90")[$90$]], [#math.equation(block: false, alt: "69")[$69$]], [#math.equation(block: false, alt: "54")[$54$]], [#math.equation(block: false, alt: "42")[$42$]], [#math.equation(block: false, alt: "33")[$33$]], [#math.equation(block: false, alt: "25")[$25$]], [#math.equation(block: false, alt: "19")[$19$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "13")[$13$]], )) + Graph the data and use your calculator's exponential regression feature to fit a curve to them. + Write your equation in the form #math.equation(block: false, alt: "G open parenthesis t close parenthesis equals G sub 0 e to the power k t")[$G ( t ) = G_(0) e^(k t)$]. + Calculate the half-life of iodine-128. + #figure(figph[decay fit on data], alt: "decay fit on data", caption: none) #linebreak() #math.equation(block: false, alt: "y equals 116 open parenthesis 0.975 close parenthesis to the power t")[$y = 116 ( 0.975 )^(t)$] + #math.equation(block: false, alt: "G open parenthesis t close parenthesis equals 116 e to the power minus 0.025 t")[$G ( t ) = 116 e^(− 0.025 t)$] + #math.equation(block: false, alt: "28")[$28$] minutes The table shows the count rate for sodium-24 registered by a Geiger counter as a function of time. (Source: Hunt and Sykes, 1984) #figure(table( columns: 11, align: left, inset: 6pt, table.header([Time (min)], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "60")[$60$]], [#math.equation(block: false, alt: "70")[$70$]], [#math.equation(block: false, alt: "80")[$80$]], [#math.equation(block: false, alt: "90")[$90$]]), [Counts/sec], [#math.equation(block: false, alt: "180")[$180$]], [#math.equation(block: false, alt: "112")[$112$]], [#math.equation(block: false, alt: "71")[$71$]], [#math.equation(block: false, alt: "45")[$45$]], [#math.equation(block: false, alt: "28")[$28$]], [#math.equation(block: false, alt: "18")[$18$]], [#math.equation(block: false, alt: "11")[$11$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "3")[$3$]], )) + Graph the data and use your calculator's exponential regression feature to fit a curve to them. + Write your equation in the form #math.equation(block: false, alt: "G open parenthesis t close parenthesis equals G sub 0 e to the power k t")[$G ( t ) = G_(0) e^(k t)$]. + Calculate the half-life of sodium-24.