## College Algebra (11th Edition)

$n=-\dfrac{\log\left(\dfrac{A-Pr}{A}\right)}{\log(1+r)}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $A=\dfrac{Pr}{1-(1+r)^{-n}} ,$ for $n ,$ use the properties of equality to isolate the base that uses the needed variable. Then take the logarithm of both sides. Use the properties of logarithms and of equality to isolate the needed variable. $\bf{\text{Solution Details:}}$ Since $\dfrac{a}{b}=\dfrac{c}{d}$ implies $ad=bc$ or sometimes referred to as cross-multiplication, the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{A}{1}=\dfrac{Pr}{1-(1+r)^{-n}} \\\\ A(1-(1+r)^{-n})=1(Pr) \\\\ A(1-(1+r)^{-n})=Pr .\end{array} Using the properties of equality to isolate the base that uses $n ,$ the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{A(1-(1+r)^{-n})}{A}=\dfrac{Pr}{A} \\\\ 1-(1+r)^{-n}=\dfrac{Pr}{A} \\\\ 1-\dfrac{Pr}{A}=(1+r)^{-n} \\\\ (1+r)^{-n}=1-\dfrac{Pr}{A} \\\\ (1+r)^{-n}=\dfrac{A}{A}-\dfrac{Pr}{A} \\\\ (1+r)^{-n}=\dfrac{A-Pr}{A} .\end{array} Taking the logarithm of both sides, the equation above is equivalent to \begin{array}{l}\require{cancel} \log(1+r)^{-n}=\log\dfrac{A-Pr}{A} .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the expression above is equivalent \begin{array}{l}\require{cancel} -n\log(1+r)=\log\left(\dfrac{A-Pr}{A}\right) \\\\ \dfrac{-n\log(1+r)}{-\log(1+r)}=\dfrac{\log\left(\dfrac{A-Pr}{A}\right)}{-\log(1+r)} \\\\ n=-\dfrac{\log\left(\dfrac{A-Pr}{A}\right)}{\log(1+r)} .\end{array}
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}