## College Algebra (11th Edition)

$t=\dfrac{\log A-\log P}{n\log\left( 1+\dfrac{r}{n} \right)}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $A=P\left( 1+\dfrac{r}{n} \right)^{tn} ,$ in terms of $t ,$ use the properties of equality to isolate the needed variable. Then take the logarithm of both sides. Use the properties of logarithms and of equality to isolate the needed variable. $\bf{\text{Solution Details:}}$ Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} \dfrac{A}{P}=\dfrac{P\left( 1+\dfrac{r}{n} \right)^{tn}}{P} \\\\ \dfrac{A}{P}=\left( 1+\dfrac{r}{n} \right)^{tn} .\end{array} Taking the logarithm of both sides results to \begin{array}{l}\require{cancel} \log\dfrac{A}{P}=\log\left( 1+\dfrac{r}{n} \right)^{tn} .\end{array} Using the Power Rule of Logarithms, which is given by $\log_b x^y=y\log_bx,$ the equation above is equivalent \begin{array}{l}\require{cancel} \log\dfrac{A}{P}=(tn)\log\left( 1+\dfrac{r}{n} \right) .\end{array} Using the Quotient Rule of Logarithms, which is given by $\log_b \dfrac{x}{y}=\log_bx-\log_by,$ the equation above is equivalent \begin{array}{l}\require{cancel} \log A-\log P=(tn)\log\left( 1+\dfrac{r}{n} \right) .\end{array} Using the properties of equality to isolate the needed variable results to \begin{array}{l}\require{cancel} \dfrac{\log A-\log P}{n\log\left( 1+\dfrac{r}{n} \right)}=\dfrac{(tn)\log\left( 1+\dfrac{r}{n} \right)}{n\log\left( 1+\dfrac{r}{n} \right)} \\\\ \dfrac{\log A-\log P}{n\log\left( 1+\dfrac{r}{n} \right)}=t \\\\ t=\dfrac{\log A-\log P}{n\log\left( 1+\dfrac{r}{n} \right)} .\end{array}