#set document(title: "10.5 Solve Exponential and Logarithmic Equations", 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")) == 10.5#h(0.6em)Solve Exponential and Logarithmic Equations #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Before you get started, take this readiness quiz. Solve: #math.equation(block: false, alt: "x squared equals 16 .")[$x^(2) = 16 .$] #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "x equals 4 , x equals minus 4")[$x = 4 , #h(0.5em) x = − 4$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "x squared minus 5 x plus 6 equals 0 .")[$x^(2) − 5 x + 6 = 0 .$] #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "x equals 2 , x equals 3")[$x = 2 , #h(0.5em) x = 3$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "x open parenthesis x plus 6 close parenthesis equals 2 x plus 5 .")[$x ( x + 6 ) = 2 x + 5 .$] #linebreak() If you missed this problem, review . #solutionbox[ #math.equation(block: true, alt: "x equals minus 5 , x equals 1")[$x = − 5 , #h(0.2em) x = 1$] ] ] === Solve Logarithmic Equations Using the Properties of Logarithms In the section on logarithmic functions, we solved some equations by rewriting the equation in exponential form. Now that we have the properties of logarithms, we have additional methods we can use to solve logarithmic equations. If our equation has two logarithms we can use a property that says that if #math.equation(block: false, alt: "log sub a M equals log sub a N")[$"log"_(a) M = "log"_(a) N$] then it is true that #math.equation(block: false, alt: "M equals N .")[$M = N .$] This is the #strong[One-to-One Property of Logarithmic Equations]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[One-to-One Property of Logarithmic Equations] For #math.equation(block: false, alt: "M greater than 0 , N greater than 0 , a > 0 ,")[$M > 0 , N > 0 , #h(0.2em) "a" ">" 0 ,$] and #math.equation(block: false, alt: "a not equal to 1")[$"a" ≠ 1$] is any real number: #math.equation(block: true, alt: "If log sub a M equals log sub a N , then M equals N .")[$"If" #h(0.2em) "log"_(a) M = "log"_(a) N , #h(0.2em) "then" #h(0.2em) M = N .$] ] To use this property, we must be certain that both sides of the equation are written with the same base. Remember that logarithms are defined only for positive real numbers. Check your results in the original equation. You may have obtained a result that gives a logarithm of zero or a negative number. #examplebox("Example 1")[][ Solve: #math.equation(block: false, alt: "2 log sub 5 x equals log sub 5 81 .")[$2 "log"_(5) x = "log"_(5) 81 .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "2 log sub 5 x equals log sub 5 81")[$#h(0.3em) 2 #h(0.2em) log_(5) x #h(0.2em) = #h(0.2em) log_(5) 81$]]), [Use the Power Property.], [#math.equation(block: false, alt: "log sub 5 x squared equals log sub 5 81")[$#h(0.2em) log_(5) x^(2) #h(0.2em) = #h(0.2em) log_(5) 81$]], [Use the One-to-One Property, if #math.equation(block: false, alt: "log sub a M equals log sub a N")[$log_(a) M = log_(a) N$], then #math.equation(block: false, alt: "M equals N")[$M = N$]], [#math.equation(block: false, alt: "x squared equals 81")[$#h(2.1em) x^(2) #h(0.2em) = #h(0.2em) 81$].], [Solve using the Square Root Property.], [#math.equation(block: false, alt: "x equals ± 9")[$#h(2.8em) x #h(0.2em) = #h(0.2em) "±" 9$]], [We eliminate #math.equation(block: false, alt: "x equals −9")[$x = −9$] as we cannot take the logarithm of a negative number.], [#math.equation(block: false, alt: "x equals 9 , x equals −9")[$#h(2.9em) x = 9 , #h(0.2em) cancel(x = −9)$]], [Check.], [], [#math.equation(block: false, alt: "x equals 9, 2 log sub 5 x, equals, log sub 5 81; 2 log sub 5 9, equals ?, log sub 5 81; log sub 5 9 squared, equals ?, log sub 5 81; log sub 5 81, equals, log sub 5 81 ✓")[$x = 9 & 2 log_(5) x & = & log_(5) 81 \ & 2 log_(5) 9 & limits(=)^(?) & log_(5) 81 \ & log_(5) 9^(2) & limits(=)^(?) & log_(5) 81 \ & log_(5) 81 & = & log_(5) 81 ✓$]], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "2 log sub 3 x equals log sub 3 36")[$2 "log"_(3) x = "log"_(3) 36$] #solutionbox[ #math.equation(block: true, alt: "x equals 6")[$x = 6$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "3 log x equals log 64")[$3 "log" #h(0.2em) x = "log" 64$] #solutionbox[ #math.equation(block: true, alt: "x equals 4")[$x = 4$] ] ] Another strategy to use to solve logarithmic equations is to condense sums or differences into a single logarithm. #examplebox("Example 2")[][ Solve: #math.equation(block: false, alt: "log sub 3 x plus log sub 3 open parenthesis x minus 8 close parenthesis equals 2 .")[$"log"_(3) x + "log"_(3) ( x − 8 ) = 2 .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "log sub 3 x plus log sub 3 open parenthesis x minus 8 close parenthesis equals 2")[$log_(3) x + log_(3) ( x − 8 ) #h(0.2em) = #h(0.2em) 2$]]), [Use the Product Property, #math.equation(block: false, alt: "log sub a M plus log sub a N equals log sub a M times N")[$log_(a) M + log_(a) N = log_(a) M ⋅ N$].], [#math.equation(block: false, alt: "log sub 3 x open parenthesis x minus 8 close parenthesis equals 2")[$#h(3em) log_(3) x ( x − 8 ) #h(0.2em) = #h(0.2em) 2$]], [Rewrite in exponential form.], [#math.equation(block: false, alt: "3 squared equals x open parenthesis x minus 8 close parenthesis")[$#h(7.5em) 3^(2) #h(0.2em) = #h(0.2em) x ( x − 8 )$]], [Simplify.], [#math.equation(block: false, alt: "9 equals x squared minus 8 x")[$#h(8em) 9 #h(0.2em) = #h(0.2em) x^(2) − 8 x$]], [Subtract 9 from each side.], [#math.equation(block: false, alt: "0 equals x squared minus 8 x minus 9")[$#h(8em) 0 #h(0.2em) = #h(0.2em) x^(2) − 8 x − 9$]], [Factor.], [#math.equation(block: false, alt: "0 equals open parenthesis x minus 9 close parenthesis open parenthesis x plus 1 close parenthesis")[$#h(8em) 0 #h(0.2em) = #h(0.2em) ( x − 9 ) ( x + 1 )$]], [Use the Zero-Product Property.], [#math.equation(block: false, alt: "x minus 9 equals 0 , x plus 1 equals 0")[$#h(6.5em) x − 9 #h(0.2em) = #h(0.2em) 0 , #h(1em) x + 1 = 0$]], [Solve each equation.], [#math.equation(block: false, alt: "x equals 9 , x equals −1")[$#h(8.5em) x = 9 , #h(2.5em) cancel(x = −1)$]], [Check.], [], [#math.equation(block: false, alt: "x equals −1, log sub 3 x plus log sub 3 open parenthesis x minus 8 close parenthesis equals 2; log sub 3 open parenthesis −1 close parenthesis plus log sub 3 open parenthesis −1 −8 close parenthesis equals ? 2")[$x = −1 & log_(3) x + log_(3) ( x − 8 ) #h(0.2em) = #h(0.2em) 2 \ & log_(3) ( −1 ) + log_(3) ( −1 −8 ) #h(0.2em) limits(=)^(?) #h(0.2em) 2$]], [], [We cannot take the log of a negative number.], [], [#math.equation(block: false, alt: "x equals 9, log sub 3 x plus log sub 3 open parenthesis x minus 8 close parenthesis equals 2 ; log sub 3 9 plus log sub 3 open parenthesis 9 minus 8 close parenthesis equals ? 2 ; 2 plus 0 equals ? 2 ; 2 equals 2 ✓")[$x = 9 & log_(3) x + log_(3) ( x − 8 ) #h(0.2em) = #h(0.2em) 2 #h(0.8em) \ & log_(3) 9 + log_(3) ( 9 − 8 ) #h(0.2em) limits(=)^(?) #h(0.2em) 2 #h(0.8em) \ & 2 + 0 #h(0.2em) limits(=)^(?) #h(0.2em) 2 #h(0.8em) \ & 2 #h(0.2em) = #h(0.2em) 2 ✓$]], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "log sub 2 x plus log sub 2 open parenthesis x minus 2 close parenthesis equals 3")[$"log"_(2) x + "log"_(2) ( x − 2 ) = 3$] #solutionbox[ #math.equation(block: true, alt: "x equals 4")[$x = 4$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "log sub 2 x plus log sub 2 open parenthesis x minus 6 close parenthesis equals 4")[$"log"_(2) x + "log"_(2) ( x − 6 ) = 4$] #solutionbox[ #math.equation(block: true, alt: "x equals 8")[$x = 8$] ] ] When there are logarithms on both sides, we condense each side into a single logarithm. Remember to use the Power Property as needed. #examplebox("Example 3")[][ Solve: #math.equation(block: false, alt: "log sub 4 open parenthesis x plus 6 close parenthesis minus log sub 4 open parenthesis 2 x plus 5 close parenthesis equals − log sub 4 x .")[$"log"_(4) ( x + 6 ) − "log"_(4) ( 2 x + 5 ) = "−" "log"_(4) x .$] #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "log sub 4 open parenthesis x plus 6 close parenthesis minus log sub 4 open parenthesis 2 x plus 5 close parenthesis equals − log sub 4 x")[$"log"_(4) ( x + 6 ) − "log"_(4) ( 2 x + 5 ) = "−" "log"_(4) x$]]), [Use the Quotient Property on the left side and the Power Property on the right.], [#math.equation(block: false, alt: "log sub 4 open parenthesis the fraction x plus 6 over 2 x plus 5 close parenthesis equals log sub 4 x to the power −1")[$"log"_(4) ( frac(x + 6, 2 x + 5) ) = "log"_(4) x^(−1)$]], [Rewrite #math.equation(block: false, alt: "x to the power −1 equals the fraction 1 over x")[$x^(−1) = frac(1, x)$].], [#math.equation(block: false, alt: "log sub 4 open parenthesis the fraction x plus 6 over 2 x plus 5 close parenthesis equals log sub 4 the fraction 1 over x")[$"log"_(4) ( frac(x + 6, 2 x + 5) ) = "log"_(4) frac(1, x) #h(0.3em)$]], [Use the One-to-One Property, if #math.equation(block: false, alt: "log sub a M equals log sub a N")[$"log"_(a) M = "log"_(a) N$], then #math.equation(block: false, alt: "M equals N")[$M = N$].], [#math.equation(block: false, alt: "the fraction x plus 6 over 2 x plus 5 equals the fraction 1 over x")[$frac(x + 6, 2 x + 5) = frac(1, x) #h(2.2em)$]], [Solve the rational equation.], [#math.equation(block: false, alt: "x open parenthesis x plus 6 close parenthesis equals 2 x plus 5")[$x ( x + 6 ) = 2 x + 5$]], [Distribute.], [#math.equation(block: false, alt: "x squared plus 6 x equals 2 x plus 5")[$x^(2) + 6 x = 2 x + 5$]], [Write in standard form.], [#math.equation(block: false, alt: "x squared plus 4 x minus 5 equals 0")[$x^(2) + 4 x − 5 = 0 #h(2.2em)$]], [Factor.], [#math.equation(block: false, alt: "open parenthesis x plus 5 close parenthesis open parenthesis x minus 1 close parenthesis equals 0")[$( x + 5 ) ( x − 1 ) = 0 #h(2.2em)$]], [Use the Zero-Product Property.], [#math.equation(block: false, alt: "x plus 5 equals 0 , x minus 1 equals 0")[$#h(17em) x + 5 = 0 , #h(2em) x − 1 = 0$]], [Solve each equation.], [#math.equation(block: false, alt: "x equals −5 , x equals 1")[$#h(22.9em) cancel(x = −5) , #h(2.8em) x = 1$]], [Check.], [], [We leave the check for you.], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "log open parenthesis x plus 2 close parenthesis minus log open parenthesis 4 x plus 3 close parenthesis equals − log x .")[$"log" ( x + 2 ) − "log" ( 4 x + 3 ) = "−" "log" #h(0.2em) x .$] #solutionbox[ #math.equation(block: true, alt: "x equals 3")[$x = 3$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve: #math.equation(block: false, alt: "log open parenthesis x minus 2 close parenthesis minus log open parenthesis 4 x plus 16 close parenthesis equals log the fraction 1 over x .")[$"log" ( x − 2 ) − "log" ( 4 x + 16 ) = "log" frac(1, x) .$] #solutionbox[ #math.equation(block: true, alt: "x equals 8")[$x = 8$] ] ] === Solve Exponential Equations Using Logarithms In the section on exponential functions, we solved some equations by writing both sides of the equation with the same base. Next we wrote a new equation by setting the exponents equal. It is not always possible or convenient to write the expressions with the same base. In that case we often take the common logarithm or natural logarithm of both sides once the exponential is isolated. #examplebox("Example 4")[][ Solve #math.equation(block: false, alt: "5 to the power x equals 11 .")[$5^(x) = 11 .$] Find the exact answer and then approximate it to three decimal places. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, [], [#math.equation(block: false, alt: "5 to the power x, equals, 11")[$#h(1.3em) 5^(x) & = & 11$]], [Since the exponential is isolated, take the logarithm of both sides. #linebreak() Use the Power Property to get the #math.equation(block: false, alt: "x")[$x$] as a factor, not an exponent. #linebreak() #linebreak() Solve for #math.equation(block: false, alt: "x .")[$x .$] Find the exact answer. #linebreak() #linebreak() Approximate the answer.], [#math.equation(block: false, alt: "log 5 to the power x, equals, log 11; x log 5, equals, log 11; x, equals, the fraction log 11 over log 5; x, approximately equals, 1.490")[$"log" 5^(x) & = & "log" 11 \ x "log" 5 & = & "log" 11 \ x & = & frac("log" 11, "log" 5) \ x & ≈ & 1.490$]], [Since #math.equation(block: false, alt: "5 to the power 1 equals 5")[$5^(1) = 5$] and #math.equation(block: false, alt: "5 squared equals 25 ,")[$5^(2) = 25 ,$] does it makes sense that #math.equation(block: false, alt: "5 to the power 1.490 approximately equals 11 ?")[$5^(1.490) ≈ 11 ?$]], [], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "7 to the power x equals 43 .")[$7^(x) = 43 .$] Find the exact answer and then approximate it to three decimal places. #solutionbox[ #math.equation(block: true, alt: "x equals the fraction log 43 over log 7 approximately equals 1.933")[$x = frac("log" 43, "log" 7) ≈ 1.933$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "8 to the power x equals 98 .")[$8^(x) = 98 .$] Find the exact answer and then approximate it to three decimal places. #solutionbox[ #math.equation(block: true, alt: "x equals the fraction log 98 over log 8 approximately equals 2.205")[$x = frac("log" 98, "log" 8) ≈ 2.205$] ] ] When we take the logarithm of both sides we will get the same result whether we use the common or the natural logarithm (try using the natural log in the last example. Did you get the same result?) When the exponential has base #emph[e], we use the natural logarithm. #examplebox("Example 5")[][ Solve #math.equation(block: false, alt: "3 e to the power x plus 2 equals 24 .")[$3 e^(x + 2) = 24 .$] Find the exact answer and then approximate it to three decimal places. #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "3 e to the power x plus 2 equals 24")[$3 e^(x + 2) #h(0.2em) = #h(0.2em) 24 #h(2.1em)$]]), [Isolate the exponential by dividing both sides by 3.], [#math.equation(block: false, alt: "e to the power x plus 2 equals 8")[$e^(x + 2) #h(0.2em) = #h(0.2em) 8 #h(2.6em)$]], [Take the natural logarithm of both sides.], [#math.equation(block: false, alt: "ln e to the power x plus 2 equals ln 8")[$"ln" #h(0.2em) e^(x + 2) #h(0.2em) = #h(0.2em) "ln" 8 #h(1.8em)$]], [Use the Power Property to get the #math.equation(block: false, alt: "x")[$x$] as a factor, not an exponent.], [#math.equation(block: false, alt: "open parenthesis x plus 2 close parenthesis ln e equals ln 8")[$( x + 2 ) "ln" #h(0.2em) e #h(0.2em) = #h(0.2em) "ln" 8 #h(1.8em)$]], [Use the property #math.equation(block: false, alt: "ln e equals 1")[$"ln" #h(0.2em) e = 1$] to simplify.], [#math.equation(block: false, alt: "x plus 2 equals ln 8")[$x + 2 #h(0.2em) = #h(0.2em) "ln" 8 #h(1.8em)$]], [Solve the equation. Find the exact answer.], [#math.equation(block: false, alt: "x equals ln 8 minus 2")[$x #h(0.2em) = #h(0.2em) "ln" 8 − 2$]], [Approximate the answer.], [#math.equation(block: false, alt: "x approximately equals 0.079")[$x #h(0.2em) ≈ #h(0.2em) 0.079 #h(0.8em)$]], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "2 e to the power x minus 2 equals 18 .")[$2 e^(x − 2) = 18 .$] Find the exact answer and then approximate it to three decimal places. #solutionbox[ #math.equation(block: true, alt: "x equals ln 9 plus 2 approximately equals 4.197")[$x = "ln" 9 + 2 ≈ 4.197$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve #math.equation(block: false, alt: "5 e to the power 2 x equals 25 .")[$5 e^(2 x) = 25 .$] Find the exact answer and then approximate it to three decimal places. #solutionbox[ #math.equation(block: true, alt: "x equals the fraction ln 5 over 2 approximately equals 0.805")[$x = frac("ln" 5, 2) ≈ 0.805$] ] ] === Use Exponential Models in Applications In previous sections we were able to solve some applications that were modeled with exponential equations. Now that we have so many more options to solve these equations, we are able to solve more applications. We will again use the Compound Interest Formulas and so we list them here for reference. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Compound Interest] For a principal, #emph[P], invested at an interest rate, #emph[r], for #emph[t] years, the new balance, #emph[A] is: #math.equation(block: true, alt: "A equals P open parenthesis 1 plus the fraction r over n close parenthesis to the power n t, when compounded n times a year.; A equals P e to the power r t, when compounded continuously.")[$#h(5em) A = P attach(( 1 + frac(r, n) ), t: n t) & & & & & "when compounded" #h(0.2em) n #h(0.2em) "times a year." \ #h(5em) A = P e^(r t) & & & & & "when compounded continuously."$] ] #examplebox("Example 6")[][ Jermael’s parents put \$10,000 in investments for his college expenses on his first birthday. They hope the investments will be worth \$50,000 when he turns 18. If the interest compounds continuously, approximately what rate of growth will they need to achieve their goal? #solutionbox[ #figure(table( columns: 2, align: left, inset: 6pt, table.header([], [#math.equation(block: false, alt: "A equals $ 50,000")[$#h(4em) A = "$" 50,000$]]), [], [#math.equation(block: false, alt: "P equals $ 10,000")[$#h(4em) P = "$" 10,000$]], [Identify the variables in the formula], [#math.equation(block: false, alt: "r equals ?")[$#h(4em) r = ?$]], [], [#math.equation(block: false, alt: "t equals 17 years")[$#h(4em) t = 17 #h(0.2em) "years"$]], [], [#math.equation(block: false, alt: "A equals P e to the power r t")[$#h(4em) A = P e^(r t)$]], [Substitute the values into the formula.], [#math.equation(block: false, alt: "50,000 equals 10,000 e to the power r times 17")[$#h(1.8em) 50,000 = 10,000 e^(r · 17)$]], [Solve for #math.equation(block: false, alt: "r .")[$r .$] Divide each side by 10,000.], [#math.equation(block: false, alt: "5 equals e to the power 17 r")[$#h(4em) 5 = e^(17 r)$]], [Take the natural log of each side.], [#math.equation(block: false, alt: "ln 5 equals ln e to the power 17 r")[$#h(3.2em) "ln" 5 = "ln" #h(0.2em) e^(17 r)$]], [Use the Power Property.], [#math.equation(block: false, alt: "ln 5 equals 17 r ln e")[$#h(3.2em) "ln" 5 = 17 r "ln" #h(0.2em) e$]], [Simplify.], [#math.equation(block: false, alt: "ln 5 equals 17 r")[$#h(3.2em) "ln" 5 = 17 r$]], [Divide each side by 17.], [#math.equation(block: false, alt: "the fraction ln 5 over 17 equals r")[$#h(3.2em) frac("ln" 5, 17) = r$]], [Approximate the answer.], [#math.equation(block: false, alt: "r approximately equals 0.095")[$#h(4em) r ≈ 0.095$]], [Convert to a percentage.], [#math.equation(block: false, alt: "r approximately equals 9.5 %")[$#h(4em) r ≈ 9.5 "%"$]], [], [They need the rate of growth to be approximately #math.equation(block: false, alt: "9.5 %")[$9.5 "%"$].], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Hector invests #math.equation(block: false, alt: "$ 10,000")[$"$" 10,000$] at age 21. He hopes the investments will be worth #math.equation(block: false, alt: "$ 150,000")[$"$" 150,000$] when he turns 50. If the interest compounds continuously, approximately what rate of growth will he need to achieve his goal? #solutionbox[ #math.equation(block: true, alt: "r approximately equals 9.3 %")[$r ≈ 9.3 "%"$] ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Rachel invests #math.equation(block: false, alt: "$ 15,000")[$"$" 15,000$] at age 25. She hopes the investments will be worth #math.equation(block: false, alt: "$ 90,000")[$"$" 90,000$] when she turns 40. If the interest compounds continuously, approximately what rate of growth will she need to achieve her goal? #solutionbox[ #math.equation(block: true, alt: "r approximately equals 11.9 %")[$r ≈ 11.9 "%"$] ] ] We have seen that growth and decay are modeled by exponential functions. For growth and decay we use the formula #math.equation(block: false, alt: "A equals A sub 0 e to the power k t .")[$A = A_(0) e^(k t) .$] Exponential growth has a positive rate of growth or growth constant, #math.equation(block: false, alt: "k")[$k$], and #strong[exponential decay] has a negative rate of growth or decay constant, #emph[k]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Exponential Growth and Decay] For an original amount, #math.equation(block: false, alt: "A sub 0 ,")[$A_(0) ,$] that grows or decays at a rate, #emph[k], for a certain time, #emph[t], the final amount, #emph[A], is: #math.equation(block: true, alt: "A equals A sub 0 e to the power k t")[$A = A_(0) e^(k t)$] ] We can now solve applications that give us enough information to determine the rate of growth. We can then use that rate of growth to predict other situations. #examplebox("Example 7")[][ Researchers recorded that a certain bacteria population grew from 100 to 300 in 3 hours. At this rate of growth, how many bacteria will there be 24 hours from the start of the experiment? #solutionbox[ This problem requires two main steps. First we must find the unknown rate, #emph[k]. Then we use that value of #emph[k] to help us find the unknown number of bacteria. #figure(table( columns: 2, align: left, inset: 6pt, table.header([Identify the variables in the formula.], [#math.equation(block: false, alt: "A, equals, 300; A sub 0, equals, 100; k, equals, ?; t, equals, 3 hours; A, equals, A sub 0 e to the power k t")[$A & = & 300 \ A_(0) & = & 100 \ k & = & ? \ t & = & 3 #h(0.2em) "hours" \ A & = & A_(0) e^(k t)$]]), [Substitute the values in the formula.], [#math.equation(block: false, alt: "300 equals 100 e to the power k times 3")[$#h(0.1em) 300 = 100 e^(k · 3)$]], [Solve for #math.equation(block: false, alt: "k")[$k$]. Divide each side by 100.], [#math.equation(block: false, alt: "3 equals e to the power 3 k")[$#h(1.15em) 3 = e^(3 k)$]], [Take the natural log of each side.], [#math.equation(block: false, alt: "ln 3 equals ln e to the power 3 k")[$#h(0.4em) "ln" 3 = "ln" #h(0.2em) e^(3 k)$]], [Use the Power Property.], [#math.equation(block: false, alt: "ln 3 equals 3 k ln e")[$#h(0.4em) "ln" 3 = 3 k "ln" #h(0.2em) e$]], [Simplify.], [#math.equation(block: false, alt: "ln 3 equals 3 k")[$#h(0.4em) "ln" 3 = 3 k$]], [Divide each side by 3.], [#math.equation(block: false, alt: "the fraction ln 3 over 3 equals k")[$#h(0.3em) frac("ln" 3, 3) = k$]], [Approximate the answer.], [#math.equation(block: false, alt: "k approximately equals 0.366")[$#h(1.15em) k ≈ 0.366$]], [We use this rate of growth to predict the number of bacteria there will be in 24 hours.], [#math.equation(block: false, alt: "A, equals, ?; A sub 0, equals, 100; k, equals, the fraction ln 3 over 3; t, equals, 24 hours; A, equals, A sub 0 e to the power k t")[$A & = & ? \ A_(0) & = & 100 \ k & = & frac("ln" 3, 3) \ t & = & 24 #h(0.2em) "hours" \ A & = & A_(0) e^(k t)$]], [Substitute in the values.], [#math.equation(block: false, alt: "A equals 100 e to the power the fraction ln 3 over 3 times 24")[$A = 100 e^(frac("ln" 3, 3) · 24)$]], [Evaluate.], [#math.equation(block: false, alt: "A approximately equals 656,100")[$A ≈ 656,100$]], [], [At this rate of growth, they can expect 656,100 bacteria.], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Researchers recorded that a certain bacteria population grew from 100 to 500 in 6 hours. At this rate of growth, how many bacteria will there be 24 hours from the start of the experiment? #solutionbox[ There will be 62,500 bacteria. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Researchers recorded that a certain bacteria population declined from 700,000 to 400,000 in 5 hours after the administration of medication. At this rate of decay, how many bacteria will there be 24 hours from the start of the experiment? #solutionbox[ There will be 47,700 bacteria. ] ] Radioactive substances decay or decompose according to the exponential decay formula. The amount of time it takes for the substance to decay to half of its original amount is called the #strong[half-life] of the substance. Similar to the previous example, we can use the given information to determine the constant of decay, and then use that constant to answer other questions. #examplebox("Example 8")[][ The half-life of radium-226 is 1,590 years. How much of a 100 mg sample will be left in 500 years? #solutionbox[ This problem requires two main steps. First we must find the decay constant #emph[k]. If we start with 100-mg, at the half-life there will be 50-mg remaining. We will use this information to find #emph[k]. Then we use that value of #emph[k] to help us find the amount of sample that will be left in 500 years. #figure(table( columns: 2, align: left, inset: 6pt, table.header([Identify the variables in the formula.], [#math.equation(block: false, alt: "A, equals, 50; A sub 0, equals, 100; k, equals, ?; t, equals, 1590 years; A, equals, A sub 0 e to the power k t")[$#h(0.2em) A & = & 50 \ A_(0) & = & 100 \ k & = & ? \ t & = & 1590 "years" \ A & = & A_(0) e^(k t)$]]), [Substitute the values in the formula.], [#math.equation(block: false, alt: "50 equals 100 e to the power k times 1590")[$#h(0.88em) 50 = 100 e^(k · 1590)$]], [Solve for #math.equation(block: false, alt: "k")[$k$]. Divide each side by 100.], [#math.equation(block: false, alt: "0.5 equals e to the power 1590 k")[$#h(0.7em) 0.5 = e^(1590 k)$]], [Take the natural log of each side.], [#math.equation(block: false, alt: "ln 0.5 equals ln e to the power 1590 k")[$"ln" 0.5 = "ln" #h(0.2em) e^(1590 k)$]], [Use the Power Property.], [#math.equation(block: false, alt: "ln 0.5 equals 1590 k ln e")[$"ln" 0.5 = 1590 k "ln" #h(0.2em) e$]], [Simplify.], [#math.equation(block: false, alt: "ln 0.5 equals 1590 k")[$"ln" 0.5 = 1590 k$]], [Divide each side by 1590.], [#math.equation(block: false, alt: "the fraction ln 0.5 over 1590 equals k exact answer")[$frac("ln" 0.5, 1590) = k #h(0.2em) "exact answer"$]], [We use this rate of growth to predict the amount that will be left in 500 years.], [#math.equation(block: false, alt: "A, equals, ?; A sub 0, equals, 100; k, equals, the fraction ln 0.5 over 1590; t, equals, 500 years; A, equals, A sub 0 e to the power k t")[$#h(0.5em) A & = & ? \ A_(0) & = & 100 \ k & = & frac("ln" 0.5, 1590) \ t & = & 500 "years" \ A & = & A_(0) e^(k t)$]], [Substitute in the values.], [#math.equation(block: false, alt: "A equals 100 e to the power the fraction ln 0.5 over 1590 times 500")[$A = 100 e^(frac("ln" 0.5, 1590) · 500)$]], [Evaluate.], [#math.equation(block: false, alt: "A approximately equals 80.4 mg")[$A ≈ 80.4 #h(0.2em) "mg"$]], [], [In 500 years there would be approximately 80.4 mg remaining.], )) ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The half-life of magnesium-27 is 9.45 minutes. How much of a 10-mg sample will be left in 6 minutes? #solutionbox[ There will be 6.44 mg left. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The half-life of radioactive iodine is 60 days. How much of a 50-mg sample will be left in 40 days? #solutionbox[ There will be 31.5 mg left. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Access these online resources for additional instruction and practice with solving exponential and logarithmic equations. - #link("https://openstax.org/l/37Solvelog")[Solving Logarithmic Equations] - #link("https://openstax.org/l/37sSolvelogeqs2")[Solving Logarithm Equations] - #link("https://openstax.org/l/37Solveforrate")[Finding the rate or time in a word problem on exponential growth or decay] - #link("https://openstax.org/l/37Solveforrate2")[Finding the rate or time in a word problem on exponential growth or decay] ] === Key Concepts - #strong[One-to-One Property of Logarithmic Equations:] For #math.equation(block: false, alt: "M greater than 0 , N greater than 0 , a > 0 ,")[$M > 0 , N > 0 , #h(0.2em) a #h(0.2em) ">" #h(0.2em) 0 ,$] and #math.equation(block: false, alt: "a not equal to 1")[$"a" ≠ 1$] is any real number: #linebreak() #math.equation(block: true, alt: "If log sub a M equals log sub a N , then M equals N .")[$"If" #h(0.2em) "log"_(a) M = "log"_(a) N , #h(0.2em) "then" #h(0.2em) M = N .$] - #strong[Compound Interest:] #linebreak() For a principal, #emph[P], invested at an interest rate, #emph[r], for #emph[t] years, the new balance, #emph[A], is: #linebreak() #math.equation(block: true, alt: "A equals P open parenthesis 1 plus the fraction r over n close parenthesis to the power n t, when compounded n times a year.; A equals P e to the power r t, when compounded continuously.")[$#h(5em) A = P attach(( 1 + frac(r, n) ), t: n t) & & & & & "when compounded" #h(0.2em) n #h(0.2em) "times a year." \ #h(5em) A = P e^(r t) & & & & & "when compounded continuously."$] - #strong[Exponential Growth and Decay:] For an original amount, #math.equation(block: false, alt: "A sub 0")[$A_(0)$] that grows or decays at a rate, #emph[r], for a certain time #emph[t], the final amount, #emph[A], is #math.equation(block: false, alt: "A equals A sub 0 e to the power r t .")[$A = A_(0) e^(r t) .$] === Section Exercises ==== Practice Makes Perfect #strong[Solve Logarithmic Equations Using the Properties of Logarithms] In the following exercises, solve for #emph[x]. #math.equation(block: true, alt: "log sub 4 64 equals 2 log sub 4 x")[$"log"_(4) 64 = 2 "log"_(4) x$] #math.equation(block: true, alt: "log 49 equals 2 log x")[$"log" 49 = 2 "log" #h(0.2em) x$] #solutionbox[ #math.equation(block: true, alt: "x equals 7")[$x = 7$] ] #math.equation(block: true, alt: "3 log sub 3 x equals log sub 3 27")[$3 "log"_(3) x = "log"_(3) 27$] #math.equation(block: true, alt: "3 log sub 6 x equals log sub 6 64")[$3 "log"_(6) x = "log"_(6) 64$] #solutionbox[ #math.equation(block: true, alt: "x equals 4")[$x = 4$] ] #math.equation(block: true, alt: "log sub 5 open parenthesis 4 x minus 2 close parenthesis equals log sub 5 10")[$"log"_(5) ( 4 x − 2 ) = "log"_(5) 10$] #math.equation(block: true, alt: "log sub 3 open parenthesis x squared plus 3 close parenthesis equals log sub 3 4 x")[$"log"_(3) ( x^(2) + 3 ) = "log"_(3) 4 x$] #solutionbox[ #math.equation(block: true, alt: "x equals 1 ,")[$x = 1 ,$] #math.equation(block: true, alt: "x equals 3")[$x = 3$] ] #math.equation(block: true, alt: "log sub 3 x plus log sub 3 x equals 2")[$"log"_(3) x + "log"_(3) x = 2$] #math.equation(block: true, alt: "log sub 4 x plus log sub 4 x equals 3")[$"log"_(4) x + "log"_(4) x = 3$] #solutionbox[ #math.equation(block: true, alt: "x equals 8")[$x = 8$] ] #math.equation(block: true, alt: "log sub 2 x plus log sub 2 open parenthesis x minus 3 close parenthesis equals 2")[$"log"_(2) x + "log"_(2) ( x − 3 ) = 2$] #math.equation(block: true, alt: "log sub 3 x plus log sub 3 open parenthesis x plus 6 close parenthesis equals 3")[$"log"_(3) x + "log"_(3) ( x + 6 ) = 3$] #solutionbox[ #math.equation(block: true, alt: "x equals 3")[$x = 3$] ] #math.equation(block: true, alt: "log x plus log open parenthesis x plus 3 close parenthesis equals 1")[$"log" #h(0.2em) x + "log" ( x + 3 ) = 1$] #math.equation(block: true, alt: "log x plus log open parenthesis x minus 15 close parenthesis equals 2")[$"log" #h(0.2em) x + "log" ( x − 15 ) = 2$] #solutionbox[ #math.equation(block: true, alt: "x equals 20")[$x = 20$] ] #math.equation(block: true, alt: "log open parenthesis x plus 4 close parenthesis minus log open parenthesis 5 x plus 12 close parenthesis equals − log x")[$"log" ( x + 4 ) − "log" ( 5 x + 12 ) = "−" "log" #h(0.2em) x$] #math.equation(block: true, alt: "log open parenthesis x minus 1 close parenthesis minus log open parenthesis x plus 3 close parenthesis equals log the fraction 1 over x")[$"log" ( x − 1 ) − "log" ( x + 3 ) = "log" frac(1, x)$] #solutionbox[ #math.equation(block: true, alt: "x equals 3")[$x = 3$] ] #math.equation(block: true, alt: "log sub 5 open parenthesis x plus 3 close parenthesis plus log sub 5 open parenthesis x minus 6 close parenthesis equals log sub 5 10")[$"log"_(5) ( x + 3 ) + "log"_(5) ( x − 6 ) = "log"_(5) 10$] #math.equation(block: true, alt: "log sub 5 open parenthesis x plus 1 close parenthesis plus log sub 5 open parenthesis x minus 5 close parenthesis equals log sub 5 7")[$"log"_(5) ( x + 1 ) + "log"_(5) ( x − 5 ) = "log"_(5) 7$] #solutionbox[ #math.equation(block: true, alt: "x equals 6")[$x = 6$] ] #math.equation(block: true, alt: "log sub 3 open parenthesis 2 x minus 1 close parenthesis equals log sub 3 open parenthesis x plus 3 close parenthesis plus log sub 3 3")[$"log"_(3) ( 2 x − 1 ) = "log"_(3) ( x + 3 ) + "log"_(3) 3$] #math.equation(block: true, alt: "log open parenthesis 5 x plus 1 close parenthesis equals log open parenthesis x plus 3 close parenthesis plus log 2")[$"log" ( 5 x + 1 ) = "log" ( x + 3 ) + "log" 2$] #solutionbox[ #math.equation(block: true, alt: "x equals the fraction 5 over 3")[$x = frac(5, 3)$] ] #strong[Solve Exponential Equations Using Logarithms] In the following exercises, solve each exponential equation. Find the exact answer and then approximate it to three decimal places. #math.equation(block: true, alt: "3 to the power x equals 89")[$3^(x) = 89$] #math.equation(block: true, alt: "2 to the power x equals 74")[$2^(x) = 74$] #solutionbox[ #math.equation(block: true, alt: "x equals the fraction log 74 over log 2 approximately equals 6.209")[$x = frac("log" 74, "log" 2) ≈ 6.209$] ] #math.equation(block: true, alt: "5 to the power x equals 110")[$5^(x) = 110$] #math.equation(block: true, alt: "4 to the power x equals 112")[$4^(x) = 112$] #solutionbox[ #math.equation(block: true, alt: "x equals the fraction log 112 over log 4 approximately equals 3.404")[$x = frac("log" 112, "log" 4) ≈ 3.404$] ] #math.equation(block: true, alt: "e to the power x equals 16")[$e^(x) = 16$] #math.equation(block: true, alt: "e to the power x equals 8")[$e^(x) = 8$] #solutionbox[ #math.equation(block: true, alt: "x equals ln 8 approximately equals 2.079")[$x = "ln" 8 ≈ 2.079$] ] #math.equation(block: true, alt: "open parenthesis the fraction 1 over 2 close parenthesis to the power x equals 6")[$attach(( frac(1, 2) ), t: x) = 6$] #math.equation(block: true, alt: "open parenthesis the fraction 1 over 3 close parenthesis to the power x equals 8")[$attach(( frac(1, 3) ), t: x) = 8$] #solutionbox[ #math.equation(block: true, alt: "x equals the fraction log 8 over log the fraction 1 over 3 approximately equals − 1.893")[$x = frac("log" 8, "log" frac(1, 3)) ≈ "−" 1.893$] ] #math.equation(block: true, alt: "4 e to the power x plus 1 equals 16")[$4 e^(x + 1) = 16$] #math.equation(block: true, alt: "3 e to the power x plus 2 equals 9")[$3 e^(x + 2) = 9$] #solutionbox[ #math.equation(block: true, alt: "x equals ln 3 minus 2 approximately equals − 0.901")[$x = "ln" 3 − 2 ≈ "−" 0.901$] ] #math.equation(block: true, alt: "6 e to the power 2 x equals 24")[$6 e^(2 x) = 24$] #math.equation(block: true, alt: "2 e to the power 3 x equals 32")[$2 e^(3 x) = 32$] #solutionbox[ #math.equation(block: true, alt: "x equals the fraction ln 16 over 3 approximately equals 0.924")[$x = frac("ln" 16, 3) ≈ 0.924$] ] #math.equation(block: true, alt: "the fraction 1 over 4 e to the power x equals 3")[$frac(1, 4) e^(x) = 3$] #math.equation(block: true, alt: "the fraction 1 over 3 e to the power x equals 2")[$frac(1, 3) e^(x) = 2$] #solutionbox[ #math.equation(block: true, alt: "x equals ln 6 approximately equals 1.792")[$x = "ln" 6 ≈ 1.792$] ] #math.equation(block: true, alt: "e to the power x plus 1 plus 2 equals 16")[$e^(x + 1) + 2 = 16$] #math.equation(block: true, alt: "e to the power x minus 1 plus 4 equals 12")[$e^(x − 1) + 4 = 12$] #solutionbox[ #math.equation(block: true, alt: "x equals ln 8 plus 1 approximately equals 3.079")[$x = "ln" 8 + 1 ≈ 3.079$] ] In the following exercises, solve each equation. #math.equation(block: true, alt: "3 to the power 3 x plus 1 equals 81")[$3^(3 x + 1) = 81$] #math.equation(block: true, alt: "6 to the power 4 x minus 17 equals 216")[$6^(4 x − 17) = 216$] #solutionbox[ #math.equation(block: true, alt: "x equals 5")[$x = 5$] ] #math.equation(block: true, alt: "the fraction e to the power x squared over e to the power 14 equals e to the power 5 x")[$frac(e^(x^(2)), e^(14)) = e^(5 x)$] #math.equation(block: true, alt: "the fraction e to the power x squared over e to the power x equals e to the power 20")[$frac(e^(x^(2)), e^(x)) = e^(20)$] #solutionbox[ #math.equation(block: true, alt: "x equals −4 , x equals 5")[$x = −4 , x = 5$] ] #math.equation(block: true, alt: "log sub a 64 equals 2")[$"log"_(a) 64 = 2$] #math.equation(block: true, alt: "log sub a 81 equals 4")[$"log"_(a) 81 = 4$] #solutionbox[ #math.equation(block: true, alt: "a equals 3")[$a = 3$] ] #math.equation(block: true, alt: "ln x equals −8")[$"ln" #h(0.2em) x = −8$] #math.equation(block: true, alt: "ln x equals 9")[$"ln" #h(0.2em) x = 9$] #solutionbox[ #math.equation(block: true, alt: "x equals e to the power 9")[$x = e^(9)$] ] #math.equation(block: true, alt: "log sub 5 open parenthesis 3 x minus 8 close parenthesis equals 2")[$"log"_(5) ( 3 x − 8 ) = 2$] #math.equation(block: true, alt: "log sub 4 open parenthesis 7 x plus 15 close parenthesis equals 3")[$"log"_(4) ( 7 x + 15 ) = 3$] #solutionbox[ #math.equation(block: true, alt: "x equals 7")[$x = 7$] ] #math.equation(block: true, alt: "ln e to the power 5 x equals 30")[$"ln" #h(0.2em) e^(5 x) = 30$] #math.equation(block: true, alt: "ln e to the power 6 x equals 18")[$"ln" #h(0.2em) e^(6 x) = 18$] #solutionbox[ #math.equation(block: true, alt: "x equals 3")[$x = 3$] ] #math.equation(block: true, alt: "3 log x equals log 125")[$3 "log" #h(0.2em) x = "log" 125$] #math.equation(block: true, alt: "7 log sub 3 x equals log sub 3 128")[$7 "log"_(3) x = "log"_(3) 128$] #solutionbox[ #math.equation(block: true, alt: "x equals 2")[$x = 2$] ] #math.equation(block: true, alt: "log sub 6 x plus log sub 6 open parenthesis x minus 5 close parenthesis equals log sub 6 24")[$"log"_(6) x + "log"_(6) ( x − 5 ) = "log"_(6) 24$] #math.equation(block: true, alt: "log sub 9 x plus log sub 9 open parenthesis x minus 4 close parenthesis equals log sub 9 12")[$"log"_(9) x + "log"_(9) ( x − 4 ) = "log"_(9) 12$] #solutionbox[ #math.equation(block: true, alt: "x equals 6")[$x = 6$] ] #math.equation(block: true, alt: "log sub 2 open parenthesis x plus 2 close parenthesis minus log sub 2 open parenthesis 2 x plus 9 close parenthesis equals − log sub 2 x")[$"log"_(2) ( x + 2 ) − "log"_(2) ( 2 x + 9 ) = "−" "log"_(2) x$] #math.equation(block: true, alt: "log sub 6 open parenthesis x plus 1 close parenthesis minus log sub 6 open parenthesis 4 x plus 10 close parenthesis equals log sub 6 the fraction 1 over x")[$"log"_(6) ( x + 1 ) − "log"_(6) ( 4 x + 10 ) = "log"_(6) frac(1, x)$] #solutionbox[ #math.equation(block: true, alt: "x equals 5")[$x = 5$] ] In the following exercises, solve for #emph[x], giving an exact answer as well as an approximation to three decimal places. #math.equation(block: true, alt: "6 to the power x equals 91")[$6^(x) = 91$] #math.equation(block: true, alt: "open parenthesis the fraction 1 over 2 close parenthesis to the power x equals 10")[$attach(( frac(1, 2) ), t: x) = 10$] #solutionbox[ #math.equation(block: true, alt: "x equals the fraction log 10 over log the fraction 1 over 2 approximately equals − 3.322")[$x = frac("log" 10, "log" frac(1, 2)) ≈ "−" 3.322$] ] #math.equation(block: true, alt: "7 e to the power x minus 3 equals 35")[$7 e^(x − 3) = 35$] #math.equation(block: true, alt: "8 e to the power x plus 5 equals 56")[$8 e^(x + 5) = 56$] #solutionbox[ #math.equation(block: true, alt: "x equals ln 7 minus 5 approximately equals − 3.054")[$x = "ln" 7 − 5 ≈ "−" 3.054$] ] #strong[Use Exponential Models in Applications] In the following exercises, solve. Sung Lee invests #math.equation(block: false, alt: "$ 5,000")[$"$" 5,000$] at age 18. He hopes the investments will be worth #math.equation(block: false, alt: "$ 10,000")[$"$" 10,000$] when he turns 25. If the interest compounds continuously, approximately what rate of growth will he need to achieve his goal? Is that a reasonable expectation? Alice invests #math.equation(block: false, alt: "$ 15,000")[$"$" 15,000$] at age 30 from the signing bonus of her new job. She hopes the investments will be worth #math.equation(block: false, alt: "$ 30,000")[$"$" 30,000$] when she turns 40. If the interest compounds continuously, approximately what rate of growth will she need to achieve her goal? #solutionbox[ #math.equation(block: true, alt: "6.9 %")[$6.9 "%"$] ] Coralee invests #math.equation(block: false, alt: "$ 5,000")[$"$" 5,000$] in an account that compounds interest monthly and earns #math.equation(block: false, alt: "7 % .")[$7 "%" .$] How long will it take for her money to double? Simone invests #math.equation(block: false, alt: "$ 8,000")[$"$" 8,000$] in an account that compounds interest quarterly and earns #math.equation(block: false, alt: "5 % .")[$5 "%" .$] How long will it take for his money to double? #solutionbox[ 13.9 years ] Researchers recorded that a certain bacteria population declined from 100,000 to 100 in 24 hours. At this rate of decay, how many bacteria will there be in 16 hours? Researchers recorded that a certain bacteria population declined from 800,000 to 500,000 in 6 hours after the administration of medication. At this rate of decay, how many bacteria will there be in 24 hours? #solutionbox[ 122,070 bacteria ] A virus takes 6 days to double its original population #math.equation(block: false, alt: "open parenthesis A equals 2 A sub 0 close parenthesis .")[$( A = 2 A_(0) ) .$] How long will it take to triple its population? A bacteria doubles its original population in 24 hours #math.equation(block: false, alt: "open parenthesis A equals 2 A sub 0 close parenthesis .")[$( A = 2 A_(0) ) .$] How big will its population be in 72 hours? #solutionbox[ 8 times as large as the original population ] Carbon-14 is used for archeological carbon dating. Its half-life is 5,730 years. How much of a 100-gram sample of Carbon-14 will be left in 1000 years? Radioactive technetium-99m is often used in diagnostic medicine as it has a relatively short half-life but lasts long enough to get the needed testing done on the patient. If its half-life is 6 hours, how much of the radioactive material form a 0.5 ml injection will be in the body in 24 hours? #solutionbox[ 0.03 ml ] ==== Writing Exercises Explain the method you would use to solve these equations: #math.equation(block: false, alt: "3 to the power x plus 1 equals 81 ,")[$3^(x + 1) = 81 ,$] #math.equation(block: false, alt: "3 to the power x plus 1 equals 75 .")[$3^(x + 1) = 75 .$] Does your method require logarithms for both equations? Why or why not? What is the difference between the equation for exponential growth versus the equation for exponential decay? #solutionbox[ Answers will vary. ] ==== Self Check ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. #figure(figph[This table has four rows and four columns. The first row, which serves as a header, reads I can…, Confidently, With some help, and No—I don’t get it. The first column below the header row reads solve logarithmic equations using the properties of logarithms, solve exponential equations using logarithms, and use exponential models in applications. The rest of the cells are blank.], alt: "This table has four rows and four columns. The first row, which serves as a header, reads I can…, Confidently, With some help, and No—I don’t get it. The first column below the header row reads solve logarithmic equations using the properties of logarithms, solve exponential equations using logarithms, and use exponential models in applications. The rest of the cells are blank.", caption: none) ⓑ After looking at the checklist, do you think you are well-prepared for the next section? Why or why not? === Chapter Review Exercises ==== Finding Composite and Inverse Functions #strong[Find and Evaluate Composite Functions] In the following exercises, for each pair of functions, find ⓐ (#emph[f] ∘ #emph[g])(#emph[x]), ⓑ (#emph[g] ∘ #emph[f])(#emph[x]), and ⓒ (#emph[f] · #emph[g])(#emph[x]). #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 7 x minus 2")[$f ( x ) = 7 x − 2$] and #linebreak() #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 5 x plus 1")[$g ( x ) = 5 x + 1$] #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 4 x")[$f ( x ) = 4 x$] and #linebreak() #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals x squared plus 3 x")[$g ( x ) = x^(2) + 3 x$] #solutionbox[ ⓐ #math.equation(block: false, alt: "4 x squared plus 12 x")[$4 x^(2) + 12 x$] ⓑ #math.equation(block: false, alt: "16 x squared plus 12 x")[$16 x^(2) + 12 x$] ⓒ #math.equation(block: false, alt: "4 x cubed plus 12 x squared")[$4 x^(3) + 12 x^(2)$] ] In the following exercises, evaluate the composition. For functions #linebreak() #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 x squared plus 2")[$f ( x ) = 3 x^(2) + 2$] and #linebreak() #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 4 x minus 3 ,")[$g ( x ) = 4 x − 3 ,$] find #linebreak() ⓐ #math.equation(block: false, alt: "open parenthesis f ∘ g close parenthesis open parenthesis −3 close parenthesis")[$( f ∘ g ) ( −3 )$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis g ∘ f close parenthesis open parenthesis −2 close parenthesis")[$( g ∘ f ) ( −2 )$] #linebreak() ⓒ #math.equation(block: false, alt: "open parenthesis f ∘ f close parenthesis open parenthesis −1 close parenthesis")[$( f ∘ f ) ( −1 )$] For functions #linebreak() #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 x cubed plus 5")[$f ( x ) = 2 x^(3) + 5$] and #linebreak() #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 3 x squared minus 7 ,")[$g ( x ) = 3 x^(2) − 7 ,$] find #linebreak() ⓐ #math.equation(block: false, alt: "open parenthesis f ∘ g close parenthesis open parenthesis −1 close parenthesis")[$( f ∘ g ) ( −1 )$] #linebreak() ⓑ #math.equation(block: false, alt: "open parenthesis g ∘ f close parenthesis open parenthesis −2 close parenthesis")[$( g ∘ f ) ( −2 )$] #linebreak() ⓒ #math.equation(block: false, alt: "open parenthesis g ∘ g close parenthesis open parenthesis 1 close parenthesis")[$( g ∘ g ) ( 1 )$] #solutionbox[ ⓐ #math.equation(block: false, alt: "−123")[$−123$] ⓑ 356 ⓒ 41 ] #strong[Determine Whether a Function is One-to-One] In the following exercises, for each set of ordered pairs, determine if it represents a function and if so, is the function one-to-one. #math.equation(block: true, alt: "{ open parenthesis −3 , −5 close parenthesis , open parenthesis −2 , −4 close parenthesis , open parenthesis −1 , −3 close parenthesis , open parenthesis 0 , −2 close parenthesis ,")[$\{ ( −3 , −5 ) , ( −2 , −4 ) , ( −1 , −3 ) , ( 0 , −2 ) ,$] #linebreak() #math.equation(block: true, alt: "open parenthesis −1 , −1 close parenthesis , open parenthesis −2 , 0 close parenthesis , open parenthesis −3 , 1 close parenthesis }")[$( −1 , −1 ) , ( −2 , 0 ) , ( −3 , 1 ) \}$] #math.equation(block: true, alt: "{ open parenthesis −3 , 0 close parenthesis , open parenthesis −2 , −2 close parenthesis , open parenthesis −1 , 0 close parenthesis , open parenthesis 0 , 1 close parenthesis ,")[$\{ ( −3 , 0 ) , ( −2 , −2 ) , ( −1 , 0 ) , ( 0 , 1 ) ,$] #linebreak() #math.equation(block: true, alt: "open parenthesis 1 , 2 close parenthesis , open parenthesis 2 , 1 close parenthesis , open parenthesis 3 , −1 close parenthesis }")[$( 1 , 2 ) , ( 2 , 1 ) , ( 3 , −1 ) \}$] #solutionbox[ Function; not one-to-one ] #math.equation(block: true, alt: "{ open parenthesis −3 , 3 close parenthesis , open parenthesis −2 , 1 close parenthesis , open parenthesis −1 , −1 close parenthesis , open parenthesis 0 , −3 close parenthesis ,")[$\{ ( −3 , 3 ) , ( −2 , 1 ) , ( −1 , −1 ) , ( 0 , −3 ) ,$] #linebreak() #math.equation(block: true, alt: "open parenthesis 1 , −5 close parenthesis , open parenthesis 2 , −4 close parenthesis , open parenthesis 3 , −2 close parenthesis }")[$( 1 , −5 ) , ( 2 , −4 ) , ( 3 , −2 ) \}$] In the following exercises, determine whether each graph is the graph of a function and if so, is it one-to-one. ⓐ #linebreak() #figure(figph[This figure shows a line from (negative 6, negative 2) up to (negative 1, 3) and then down from there to (6, negative 4).], alt: "This figure shows a line from (negative 6, negative 2) up to (negative 1, 3) and then down from there to (6, negative 4).", caption: none) #linebreak() ⓑ #linebreak() #figure(figph[This figure shows a line from (6, 5) down to (0, negative 1) and then down from there to (5, negative 6).], alt: "This figure shows a line from (6, 5) down to (0, negative 1) and then down from there to (5, negative 6).", caption: none) #solutionbox[ ⓐ Function; not one-to-one ⓑ Not a function ] ⓐ #linebreak() #figure(figph[This figure shows a curved line from (negative 6, negative 2) up to the origin and then continuing up from there to (6, 2).], alt: "This figure shows a curved line from (negative 6, negative 2) up to the origin and then continuing up from there to (6, 2).", caption: none) #linebreak() ⓑ #linebreak() #figure(figph[This figure shows a circle of radius 2 with center at the origin.], alt: "This figure shows a circle of radius 2 with center at the origin.", caption: none) #strong[Find the Inverse of a Function] In the following exercise, find the inverse of the function. Determine the domain and range of the inverse function. #math.equation(block: true, alt: "{ open parenthesis −3 , 10 close parenthesis , open parenthesis −2 , 5 close parenthesis , open parenthesis −1 , 2 close parenthesis , open parenthesis 0 , 1 close parenthesis }")[$\{ ( −3 , 10 ) , ( −2 , 5 ) , ( −1 , 2 ) , ( 0 , 1 ) \}$] #solutionbox[ Inverse function: #math.equation(block: false, alt: "{ open parenthesis 10 , −3 close parenthesis , open parenthesis 5 , −2 close parenthesis , open parenthesis 2 , −1 close parenthesis , open parenthesis 1 , 0 close parenthesis } .")[$\{ ( 10 , −3 ) , ( 5 , −2 ) , ( 2 , −1 ) , ( 1 , 0 ) \} .$] Domain: #math.equation(block: false, alt: "{ 1 , 2 , 5 , 10 } .")[$\{ 1 , 2 , 5 , 10 \} .$] Range: #math.equation(block: false, alt: "{ −3 , −2 , −1 , 0 } .")[$\{ −3 , −2 , −1 , 0 \} .$] ] In the following exercise, graph the inverse of the one-to-one function shown. #figure(figph[This figure shows a line segment from (negative 4, negative 2) up to (negative 2, 1) then up to (2, 2) and then up to (3, 4).], alt: "This figure shows a line segment from (negative 4, negative 2) up to (negative 2, 1) then up to (2, 2) and then up to (3, 4).", caption: none) In the following exercises, verify that the functions are inverse functions. #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 x plus 7")[$f ( x ) = 3 x + 7$] and #linebreak() #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction x minus 7 over 3")[$g ( x ) = frac(x − 7, 3)$] #solutionbox[ #math.equation(block: false, alt: "g open parenthesis f open parenthesis x close parenthesis close parenthesis equals x ,")[$g ( f ( x ) ) = x ,$] and #math.equation(block: false, alt: "f open parenthesis g open parenthesis x close parenthesis close parenthesis equals x ,")[$f ( g ( x ) ) = x ,$] so they are inverses. ] #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 x plus 9")[$f ( x ) = 2 x + 9$] and #linebreak() #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction x plus 9 over 2")[$g ( x ) = frac(x + 9, 2)$] In the following exercises, find the inverse of each function. #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 6 x minus 11")[$f ( x ) = 6 x − 11$] #solutionbox[ #math.equation(block: true, alt: "f to the power −1 open parenthesis x close parenthesis equals the fraction x plus 11 over 6")[$f^(−1) ( x ) = frac(x + 11, 6)$] ] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x cubed plus 13")[$f ( x ) = x^(3) + 13$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the fraction 1 over x plus 5")[$f ( x ) = frac(1, x + 5)$] #solutionbox[ #math.equation(block: true, alt: "f to the power −1 open parenthesis x close parenthesis equals the fraction 1 over x minus 5")[$f^(−1) ( x ) = frac(1, x) − 5$] ] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the 5th root of x minus 1")[$f ( x ) = root(5, x − 1)$] ==== Evaluate and Graph Exponential Functions #strong[Graph Exponential Functions] In the following exercises, graph each of the following functions. #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 4 to the power x")[$f ( x ) = 4^(x)$] #solutionbox[ #figure(figph[This figure shows an exponential line passing through the points (negative 1, 1 over 4), (0, 1), and (1, 4).], alt: "This figure shows an exponential line passing through the points (negative 1, 1 over 4), (0, 1), and (1, 4).", caption: none) ] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open parenthesis the fraction 1 over 5 close parenthesis to the power x")[$f ( x ) = attach(( frac(1, 5) ), t: x)$] #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals open parenthesis 0.75 close parenthesis to the power x")[$g ( x ) = attach(( 0.75 ), t: x)$] #solutionbox[ #figure(figph[This figure shows an exponential line passing through the points (negative 1, 4 over 3), (0, 1), and (1, 3 over 4).], alt: "This figure shows an exponential line passing through the points (negative 1, 4 over 3), (0, 1), and (1, 3 over 4).", caption: none) ] #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals 3 to the power x plus 2")[$g ( x ) = 3^(x + 2)$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open parenthesis 2.3 close parenthesis to the power x minus 3")[$f ( x ) = attach(( 2.3 ), t: x) − 3$] #solutionbox[ #figure(figph[This figure shows an exponential line passing through the points (negative 1, negative 59 over 23), (0, negative 2), and (1, negative7 over 10).], alt: "This figure shows an exponential line passing through the points (negative 1, negative 59 over 23), (0, negative 2), and (1, negative7 over 10).", caption: none) ] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals e to the power x plus 5")[$f ( x ) = e^(x) + 5$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals − e to the power x")[$f ( x ) = "−" e^(x)$] #solutionbox[ #figure(figph[This figure shows an exponential line passing through the points (negative 1, negative 1 over e), (0, negative 1), and (1, negative e).], alt: "This figure shows an exponential line passing through the points (negative 1, negative 1 over e), (0, negative 1), and (1, negative e).", caption: none) ] #strong[Solve Exponential Equations] In the following exercises, solve each equation. #math.equation(block: true, alt: "3 to the power 5 x minus 6 equals 81")[$3^(5 x − 6) = 81$] #math.equation(block: true, alt: "2 to the power x squared equals 16")[$2^(x^(2)) = 16$] #solutionbox[ #math.equation(block: true, alt: "x equals −2 , x equals 2")[$x = −2 , x = 2$] ] #math.equation(block: true, alt: "9 to the power x equals 27")[$9^(x) = 27$] #math.equation(block: true, alt: "5 to the power x squared plus 2 x equals the fraction 1 over 5")[$5^(x^(2) + 2 x) = frac(1, 5)$] #solutionbox[ #math.equation(block: true, alt: "x equals −1")[$x = −1$] ] #math.equation(block: true, alt: "e to the power 4 x times e to the power 7 equals e to the power 19")[$e^(4 x) · e^(7) = e^(19)$] #math.equation(block: true, alt: "the fraction e to the power x squared over e to the power 15 equals e to the power 2 x")[$frac(e^(x^(2)), e^(15)) = e^(2 x)$] #solutionbox[ #math.equation(block: true, alt: "x equals −3 , x equals 5")[$x = −3 , x = 5$] ] #strong[Use Exponential Models in Applications] In the following exercises, solve. Felix invested #math.equation(block: false, alt: "$ 12,000")[$"$" 12,000$] in a savings account. If the interest rate is #math.equation(block: false, alt: "4 %")[$4 "%"$] how much will be in the account in 12 years by each method of compounding? ⓐ compound quarterly #linebreak() ⓑ compound monthly #linebreak() ⓒ compound continuously. Sayed deposits #math.equation(block: false, alt: "$ 20,000")[$"$" 20,000$] in an investment account. What will be the value of his investment in 30 years if the investment is earning #math.equation(block: false, alt: "7 %")[$7 "%"$] per year and is compounded continuously? #solutionbox[ #math.equation(block: true, alt: "$ 163,323.40")[$"$" 163,323.40$] ] A researcher at the Center for Disease Control and Prevention is studying the growth of a bacteria. She starts her experiment with 150 of the bacteria that grows at a rate of #math.equation(block: false, alt: "15 %")[$15 "%"$] per hour. She will check on the bacteria every 24 hours. How many bacteria will he find in 24 hours? In the last five years the population of the United States has grown at a rate of #math.equation(block: false, alt: "0.7 %")[$0.7 "%"$] per year to about 318,900,000. If this rate continues, what will be the population in 5 more years? #solutionbox[ 330,259,000 ] ==== Evaluate and Graph Logarithmic Functions #strong[Convert Between Exponential and Logarithmic Form] In the following exercises, convert from exponential to logarithmic form. #math.equation(block: true, alt: "5 to the power 4 equals 625")[$5^(4) = 625$] #math.equation(block: true, alt: "10 to the power −3 equals the fraction 1 over 1,000")[$10^(−3) = frac(1, 1","000)$] #solutionbox[ #math.equation(block: true, alt: "log the fraction 1 over 1,000 equals −3")[$"log" frac(1, 1","000) = −3$] ] #math.equation(block: true, alt: "63 to the power the fraction 1 over 5 equals the 5th root of 63")[$63^(frac(1, 5)) = root(5, 63)$] #math.equation(block: true, alt: "e to the power y equals 16")[$e^(y) = 16$] #solutionbox[ #math.equation(block: true, alt: "ln 16 equals y")[$"ln" 16 = y$] ] In the following exercises, convert each logarithmic equation to exponential form. #math.equation(block: true, alt: "7 equals log sub 2 128")[$7 = "log"_(2) 128$] #math.equation(block: true, alt: "5 equals log 100,000")[$5 = "log" 100,000$] #solutionbox[ #math.equation(block: true, alt: "100000 equals 10 to the power 5")[$100000 = 10^(5)$] ] #math.equation(block: true, alt: "4 equals ln x")[$4 = "ln" #h(0.2em) x$] #strong[Evaluate Logarithmic Functions] In the following exercises, solve for #emph[x]. #math.equation(block: true, alt: "log sub x 125 equals 3")[$"log"_(x) 125 = 3$] #solutionbox[ #math.equation(block: true, alt: "x equals 5")[$x = 5$] ] #math.equation(block: true, alt: "log sub 7 x equals −2")[$"log"_(7) x = −2$] #math.equation(block: true, alt: "log sub the fraction 1 over 2 the fraction 1 over 16 equals x")[$"log"_(frac(1, 2)) frac(1, 16) = x$] #solutionbox[ #math.equation(block: true, alt: "x equals 4")[$x = 4$] ] In the following exercises, find the exact value of each logarithm without using a calculator. #math.equation(block: true, alt: "log sub 2 32")[$"log"_(2) 32$] #math.equation(block: true, alt: "log sub 8 1")[$"log"_(8) 1$] #solutionbox[ 0 ] #math.equation(block: true, alt: "log sub 3 the fraction 1 over 9")[$"log"_(3) frac(1, 9)$] #strong[Graph Logarithmic Functions] In the following exercises, graph each logarithmic function. #math.equation(block: true, alt: "y equals log sub 5 x")[$y = "log"_(5) x$] #solutionbox[ #figure(figph[This figure shows a logarithmic line passing through the points (1 over 5, negative 1), (1, 0), and (5, 1).], alt: "This figure shows a logarithmic line passing through the points (1 over 5, negative 1), (1, 0), and (5, 1).", caption: none) ] #math.equation(block: true, alt: "y equals log sub the fraction 1 over 4 x")[$y = "log"_(frac(1, 4)) x$] #math.equation(block: true, alt: "y equals log sub 0.8 x")[$y = "log"_(0.8) x$] #solutionbox[ #figure(figph[This figure shows a logarithmic line passing through the points (4 over 5, 1), (1, 0), and (5 over 4, negative 1).], alt: "This figure shows a logarithmic line passing through the points (4 over 5, 1), (1, 0), and (5 over 4, negative 1).", caption: none) ] #strong[Solve Logarithmic Equations] In the following exercises, solve each logarithmic equation. #math.equation(block: true, alt: "log sub a 36 equals 5")[$"log"_(a) 36 = 5$] #math.equation(block: true, alt: "ln x equals −3")[$"ln" #h(0.2em) x = −3$] #solutionbox[ #math.equation(block: true, alt: "x equals e to the power −3")[$x = e^(−3)$] ] #math.equation(block: true, alt: "log sub 2 open parenthesis 5 x minus 7 close parenthesis equals 3")[$"log"_(2) ( 5 x − 7 ) = 3$] #math.equation(block: true, alt: "ln e to the power 3 x equals 24")[$"ln" #h(0.2em) e^(3 x) = 24$] #solutionbox[ #math.equation(block: true, alt: "x equals 8")[$x = 8$] ] #math.equation(block: true, alt: "log open parenthesis x squared minus 21 close parenthesis equals 2")[$"log" ( x^(2) − 21 ) = 2$] #strong[Use Logarithmic Models in Applications] What is the decibel level of a train whistle with intensity #math.equation(block: false, alt: "10 to the power −3")[$10^(−3)$] watts per square inch? #solutionbox[ 90 dB ] ==== Use the Properties of Logarithms #strong[Use the Properties of Logarithms] In the following exercises, use the properties of logarithms to evaluate. ⓐ #math.equation(block: false, alt: "log sub 7 1")[$"log"_(7) 1$] ⓑ #math.equation(block: false, alt: "log sub 12 12")[$"log"_(12) 12$] ⓐ #math.equation(block: false, alt: "5 to the power log sub 5 13")[$5^("log"_(5) 13)$] ⓑ #math.equation(block: false, alt: "log sub 3 3 to the power −9")[$"log"_(3) 3^(−9)$] #solutionbox[ ⓐ 13 ⓑ #math.equation(block: false, alt: "−9")[$−9$] ] ⓐ #math.equation(block: false, alt: "10 to the power log the square root of 5")[$10^("log" sqrt(5))$] ⓑ #math.equation(block: false, alt: "log 10 to the power −3")[$"log" 10^(−3)$] ⓐ #math.equation(block: false, alt: "e to the power ln 8")[$e^("ln" 8)$] ⓑ #math.equation(block: false, alt: "ln e to the power 5")[$"ln" #h(0.2em) e^(5)$] #solutionbox[ ⓐ 8 ⓑ 5 ] In the following exercises, use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify if possible. #math.equation(block: true, alt: "log sub 4 open parenthesis 64 x y close parenthesis")[$"log"_(4) ( 64 x y )$] #math.equation(block: true, alt: "log 10,000 m")[$"log" 10,000 m$] #solutionbox[ #math.equation(block: true, alt: "4 plus log m")[$4 + "log" m$] ] In the following exercises, use the Quotient Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible. #math.equation(block: true, alt: "log sub 7 the fraction 49 over y")[$"log"_(7) frac(49, y)$] #math.equation(block: true, alt: "ln the fraction e to the power 5 over 2")[$"ln" frac(e^(5), 2)$] #solutionbox[ #math.equation(block: true, alt: "5 minus ln 2")[$5 − "ln" 2$] ] In the following exercises, use the Power Property of Logarithms to expand each logarithm. Simplify, if possible. #math.equation(block: true, alt: "log x to the power −9")[$"log" x^(−9)$] #math.equation(block: true, alt: "log sub 4 the 7th root of z")[$"log"_(4) root(7, z)$] #solutionbox[ #math.equation(block: true, alt: "the fraction 1 over 7 log sub 4 z")[$frac(1, 7) "log"_(4) z$] ] In the following exercises, use properties of logarithms to write each logarithm as a sum of logarithms. Simplify if possible. #math.equation(block: true, alt: "log sub 3 open parenthesis the square root of 4 x to the power 7 y to the power 8 close parenthesis")[$"log"_(3) ( sqrt(4) x^(7) y^(8) )$] #math.equation(block: true, alt: "log sub 5 the fraction 8 a squared b to the power 6 c over d cubed")[$log_(5) frac(8 a^(2) b^(6) c, d^(3))$] #solutionbox[ #math.equation(block: true, alt: "log sub 5 8 plus 2 log sub 5 a plus 6 log sub 5 b")[$"log"_(5) 8 + 2 "log"_(5) a + 6 "log"_(5) b$] #linebreak() #math.equation(block: true, alt: "plus log sub 5 c minus 3 log sub 5 d")[$+ #h(0.2em) "log"_(5) c − 3 "log"_(5) d$] ] #math.equation(block: true, alt: "ln the fraction the square root of 3 x squared minus y squared over z to the power 4")[$"ln" frac(sqrt(3 x^(2) − y^(2)), z^(4))$] #math.equation(block: true, alt: "log sub 6 the cube root of the fraction 7 x squared over 6 y cubed z to the power 5")[$"log"_(6) root(3, frac(7 x^(2), 6 y^(3) z^(5)))$] #solutionbox[ #math.equation(block: true, alt: "the fraction 1 over 3 open parenthesis log sub 6 7 plus 2 log sub 6 x minus 1 minus 3 log sub 6 y")[$frac(1, 3) \( "log"_(6) 7 + 2 "log"_(6) x − 1 − 3 "log"_(6) y$] #linebreak() #math.equation(block: true, alt: "minus 5 log sub 6 z close parenthesis")[$− #h(0.2em) 5 "log"_(6) z \)$] ] In the following exercises, use the Properties of Logarithms to condense the logarithm. Simplify if possible. #math.equation(block: true, alt: "log sub 2 56 minus log sub 2 7")[$"log"_(2) 56 − "log"_(2) 7$] #math.equation(block: true, alt: "3 log sub 3 x plus 7 log sub 3 y")[$3 "log"_(3) x + 7 "log"_(3) y$] #solutionbox[ #math.equation(block: true, alt: "log sub 3 x cubed y to the power 7")[$"log"_(3) x^(3) y^(7)$] ] #math.equation(block: true, alt: "log sub 5 open parenthesis x squared minus 16 close parenthesis minus 2 log sub 5 open parenthesis x plus 4 close parenthesis")[$"log"_(5) ( x^(2) − 16 ) − 2 "log"_(5) ( x + 4 )$] #math.equation(block: true, alt: "the fraction 1 over 4 log y minus 2 log open parenthesis y minus 3 close parenthesis")[$frac(1, 4) "log" y − 2 "log" ( y − 3 )$] #solutionbox[ #math.equation(block: true, alt: "log the fraction the 4th root of y over open parenthesis y minus 3 close parenthesis squared")[$"log" frac(root(4, y), attach(( y − 3 ), t: 2))$] ] #strong[Use the Change-of-Base Formula] In the following exercises, rounding to three decimal places, approximate each logarithm. #math.equation(block: true, alt: "log sub 5 97")[$"log"_(5) 97$] #math.equation(block: true, alt: "log sub the square root of 3 16")[$"log"_(sqrt(3)) 16$] #solutionbox[ 5.047 ] ==== Solve Exponential and Logarithmic Equations #strong[Solve Logarithmic Equations Using the Properties of Logarithms] In the following exercises, solve for #emph[x]. #math.equation(block: true, alt: "3 log sub 5 x equals log sub 5 216")[$3 "log"_(5) x = "log"_(5) 216$] #math.equation(block: true, alt: "log sub 2 x plus log sub 2 open parenthesis x minus 2 close parenthesis equals 3")[$"log"_(2) x + "log"_(2) ( x − 2 ) = 3$] #solutionbox[ #math.equation(block: true, alt: "x equals 4")[$x = 4$] ] #math.equation(block: true, alt: "log open parenthesis x minus 1 close parenthesis minus log open parenthesis 3 x plus 5 close parenthesis equals − log x")[$"log" ( x − 1 ) − "log" ( 3 x + 5 ) = "−" "log" #h(0.2em) x$] #math.equation(block: true, alt: "log sub 4 open parenthesis x minus 2 close parenthesis plus log sub 4 open parenthesis x plus 5 close parenthesis equals log sub 4 8")[$"log"_(4) ( x − 2 ) + "log"_(4) ( x + 5 ) = "log"_(4) 8$] #solutionbox[ #math.equation(block: true, alt: "x equals 3")[$x = 3$] ] #math.equation(block: true, alt: "ln open parenthesis 3 x minus 2 close parenthesis equals ln open parenthesis x plus 4 close parenthesis plus ln 2")[$"ln" ( 3 x − 2 ) = "ln" ( x + 4 ) + "ln" 2$] #strong[Solve Exponential Equations Using Logarithms] In the following exercises, solve each exponential equation. Find the exact answer and then approximate it to three decimal places. #math.equation(block: true, alt: "2 to the power x equals 101")[$2^(x) = 101$] #solutionbox[ #math.equation(block: true, alt: "x equals the fraction log 101 over log 2 approximately equals 6.658")[$x = frac("log" 101, "log" 2) ≈ 6.658$] ] #math.equation(block: true, alt: "e to the power x equals 23")[$e^(x) = 23$] #math.equation(block: true, alt: "open parenthesis the fraction 1 over 3 close parenthesis to the power x equals 7")[$attach(( frac(1, 3) ), t: x) = 7$] #solutionbox[ #math.equation(block: true, alt: "x equals the fraction log 7 over log the fraction 1 over 3 approximately equals − 1.771")[$x = frac("log" 7, "log" frac(1, 3)) ≈ "−" 1.771$] ] #math.equation(block: true, alt: "7 e to the power x plus 3 equals 28")[$7 e^(x + 3) = 28$] #math.equation(block: true, alt: "e to the power x minus 4 plus 8 equals 23")[$e^(x − 4) + 8 = 23$] #solutionbox[ #math.equation(block: true, alt: "x equals ln 15 plus 4 approximately equals 6.708")[$x = "ln" 15 + 4 ≈ 6.708$] ] #strong[Use Exponential Models in Applications] Jerome invests #math.equation(block: false, alt: "$ 18,000")[$"$" 18,000$] at age 17. He hopes the investments will be worth #math.equation(block: false, alt: "$ 30,000")[$"$" 30,000$] when he turns 26. If the interest compounds continuously, approximately what rate of growth will he need to achieve his goal? Is that a reasonable expectation? Elise invests #math.equation(block: false, alt: "$ 4500")[$"$" 4500$] in an account that compounds interest monthly and earns #math.equation(block: false, alt: "6 % .")[$6 "%" .$] How long will it take for her money to double? #solutionbox[ 11.6 years ] Researchers recorded that a certain bacteria population grew from 100 to 300 in 8 hours. At this rate of growth, how many bacteria will there be in 24 hours? Mouse populations can double in 8 months #math.equation(block: false, alt: "open parenthesis A equals 2 A sub 0 close parenthesis .")[$( A = 2 A_(0) ) .$] How long will it take for a mouse population to triple? #solutionbox[ 12.7 months ] The half-life of radioactive iodine is 60 days. How much of a 50 mg sample will be left in 40 days? === Practice Test For the functions, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 6 x plus 1")[$f ( x ) = 6 x + 1$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 8 x minus 3 ,")[$g ( x ) = 8 x − 3 ,$] find ⓐ #math.equation(block: false, alt: "open parenthesis f ∘ g close parenthesis open parenthesis x close parenthesis ,")[$( f ∘ g ) ( x ) ,$] ⓑ #math.equation(block: false, alt: "open parenthesis g ∘ f close parenthesis open parenthesis x close parenthesis ,")[$( g ∘ f ) ( x ) ,$] and ⓒ #math.equation(block: false, alt: "open parenthesis f times g close parenthesis open parenthesis x close parenthesis .")[$( f · g ) ( x ) .$] #solutionbox[ ⓐ #math.equation(block: false, alt: "48 x minus 17")[$48 x − 17$] ⓑ #math.equation(block: false, alt: "48 x plus 5")[$48 x + 5$] #linebreak() ⓒ #math.equation(block: false, alt: "48 x squared minus 10 x minus 3")[$48 x^(2) − 10 x − 3$] ] Determine if the following set of ordered pairs represents a function and if so, is the function one-to-one. #math.equation(block: false, alt: "{ open parenthesis −2 , 2 close parenthesis , open parenthesis −1 , −3 close parenthesis , open parenthesis 0 , 1 close parenthesis , open parenthesis 1 , −2 close parenthesis , open parenthesis 2 , −3 close parenthesis }")[$\{ ( −2 , 2 ) , ( −1 , −3 ) , ( 0 , 1 ) , ( 1 , −2 ) , ( 2 , −3 ) \}$] Determine whether each graph is the graph of a function and if so, is it one-to-one. ⓐ #linebreak() #figure(figph[This figure shows a parabola opening to the right with vertex (negative 3, 0).], alt: "This figure shows a parabola opening to the right with vertex (negative 3, 0).", caption: none) #linebreak() ⓑ #linebreak() #figure(figph[This figure shows an exponential line passing through the points (negative 1, 1 over 2), (0, 1), and (1, 2).], alt: "This figure shows an exponential line passing through the points (negative 1, 1 over 2), (0, 1), and (1, 2).", caption: none) #solutionbox[ ⓐ Not a function ⓑ One-to-one function ] Graph, on the same coordinate system, the inverse of the one-to-one function shown. #figure(figph[This figure shows a line segment passing from the point (negative 3, 3) to (negative 1, 2) to (0, negative 2) to (2, negative 4).], alt: "This figure shows a line segment passing from the point (negative 3, 3) to (negative 1, 2) to (0, negative 2) to (2, negative 4).", caption: none) Find the inverse of the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x to the power 5 minus 9 .")[$f ( x ) = x^(5) − 9 .$] #solutionbox[ #math.equation(block: true, alt: "f to the power −1 open parenthesis x close parenthesis equals the 5th root of x plus 9")[$f^(−1) ( x ) = root(5, x + 9)$] ] Graph the function #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 2 to the power x minus 3 .")[$g ( x ) = 2^(x − 3) .$] Solve the equation #math.equation(block: false, alt: "2 to the power 2 x minus 4 equals 64 .")[$2^(2 x − 4) = 64 .$] #solutionbox[ #math.equation(block: true, alt: "x equals 5")[$x = 5$] ] Solve the equation #math.equation(block: false, alt: "the fraction e to the power x squared over e to the power 4 equals e to the power 3 x .")[$frac(e^(x^(2)), e^(4)) = e^(3 x) .$] Megan invested #math.equation(block: false, alt: "$ 21,000")[$"$" 21,000$] in a savings account. If the interest rate is #math.equation(block: false, alt: "5 % ,")[$5 "%" ,$] how much will be in the account in 8 years by each method of compounding? #linebreak() ⓐ compound quarterly #linebreak() ⓑ compound monthly #linebreak() ⓒ compound continuously. #solutionbox[ ⓐ #math.equation(block: false, alt: "$ 31,250.74")[$"$" 31,250.74$] ⓑ #math.equation(block: false, alt: "$ 31,302.29")[$"$" 31,302.29$] ⓒ #math.equation(block: false, alt: "$ 31,328.32")[$"$" 31,328.32$] ] Convert the equation from exponential to logarithmic form: #math.equation(block: false, alt: "10 to the power −2 equals the fraction 1 over 100 .")[$10^(−2) = frac(1, 100) .$] Convert the equation from logarithmic equation to exponential form: #math.equation(block: false, alt: "3 equals log sub 7 343")[$3 = "log"_(7) 343$] #solutionbox[ #math.equation(block: true, alt: "343 equals 7 cubed")[$343 = 7^(3)$] ] Solve for #emph[x]: #math.equation(block: false, alt: "log sub 5 x equals −3")[$"log"_(5) x = −3$] Evaluate #math.equation(block: false, alt: "log sub 11 1 .")[$"log"_(11) 1 .$] #solutionbox[ 0 ] Evaluate #math.equation(block: false, alt: "log sub 4 the fraction 1 over 64 .")[$"log"_(4) frac(1, 64) .$] Graph the function #linebreak() #math.equation(block: false, alt: "y equals log sub 3 x .")[$y = "log"_(3) x .$] #solutionbox[ #figure(figph[This figure shows a logarithmic line passing through (1 over 3, 1), (1, 0), and (3, 1).], alt: "This figure shows a logarithmic line passing through (1 over 3, 1), (1, 0), and (3, 1).", caption: none) ] Solve for #emph[x]: #linebreak() #math.equation(block: false, alt: "log open parenthesis x squared minus 39 close parenthesis equals 1")[$"log" ( x^(2) − 39 ) = 1$] What is the decibel level of a small fan with intensity #math.equation(block: false, alt: "10 to the power −8")[$10^(−8)$] watts per square inch? #solutionbox[ 40 dB ] Evaluate each. ⓐ #math.equation(block: false, alt: "6 to the power log sub 6 17")[$6^("log"_(6) 17)$] #linebreak() ⓑ #math.equation(block: false, alt: "log sub 9 9 to the power −3")[$"log"_(9) 9^(−3)$] In the following exercises, use properties of logarithms to write each expression as a sum of logarithms, simplifying if possible. #math.equation(block: true, alt: "log sub 5 25 a b")[$"log"_(5) 25 a b$] #solutionbox[ #math.equation(block: true, alt: "2 plus log sub 5 a plus log sub 5 b")[$2 + "log"_(5) a + "log"_(5) b$] ] #math.equation(block: true, alt: "ln the fraction e to the power 12 over 8")[$"ln" frac(e^(12), 8)$] #math.equation(block: true, alt: "log sub 2 the 4th root of the fraction 5 x cubed over 16 y squared z to the power 7")[$"log"_(2) root(4, frac(5 x^(3), 16 y^(2) z^(7)))$] #solutionbox[ #math.equation(block: true, alt: "the fraction 1 over 4 open parenthesis log sub 2 5 plus 3 log sub 2 x minus 4 minus 2 log sub 2 y")[$frac(1, 4) \( "log"_(2) 5 + 3 "log"_(2) x − 4 − 2 "log"_(2) y$] #linebreak() #math.equation(block: true, alt: "minus 7 log sub 2 z close parenthesis")[$− #h(0.2em) 7 "log"_(2) z \)$] ] In the following exercises, use the Properties of Logarithms to condense the logarithm, simplifying if possible. #math.equation(block: true, alt: "5 log sub 4 x plus 3 log sub 4 y")[$5 "log"_(4) x + 3 "log"_(4) y$] #math.equation(block: true, alt: "the fraction 1 over 6 log x minus 3 log open parenthesis x plus 5 close parenthesis")[$frac(1, 6) "log" #h(0.2em) x − 3 "log" ( x + 5 )$] #solutionbox[ #math.equation(block: true, alt: "log the fraction the 6th root of x over open parenthesis x plus 5 close parenthesis cubed")[$"log" frac(root(6, x), attach(( x + 5 ), t: 3))$] ] Rounding to three decimal places, approximate #math.equation(block: false, alt: "log sub 4 73 .")[$"log"_(4) 73 .$] Solve for #emph[x]: #linebreak() #math.equation(block: false, alt: "log sub 7 open parenthesis x plus 2 close parenthesis plus log sub 7 open parenthesis x minus 3 close parenthesis equals log sub 7 24")[$"log"_(7) ( x + 2 ) + "log"_(7) ( x − 3 ) = "log"_(7) 24$] #solutionbox[ #math.equation(block: true, alt: "x equals 6")[$x = 6$] ] In the following exercises, solve each exponential equation. Find the exact answer and then approximate it to three decimal places. #math.equation(block: true, alt: "open parenthesis the fraction 1 over 5 close parenthesis to the power x equals 9")[$attach(( frac(1, 5) ), t: x) = 9$] #math.equation(block: true, alt: "5 e to the power x minus 4 equals 40")[$5 e^(x − 4) = 40$] #solutionbox[ #math.equation(block: true, alt: "x equals ln 8 plus 4 approximately equals 6.079")[$x = "ln" 8 + 4 ≈ 6.079$] ] Jacob invests \$14,000 in an account that compounds interest quarterly and earns #math.equation(block: false, alt: "4 % .")[$4 "%" .$] How long will it take for his money to double? Researchers recorded that a certain bacteria population grew from 500 to 700 in 5 hours. At this rate of growth, how many bacteria will there be in 20 hours? #solutionbox[ 1,921 bacteria ] A certain beetle population can double in 3 months #math.equation(block: false, alt: "open parenthesis A equals 2 A sub 0 close parenthesis .")[$( A = 2 A_(0) ) .$] How long will it take for that beetle population to triple?