#### Answer

$t=\dfrac{\log A-\log P}{n\log\left( 1+\dfrac{r}{n} \right)}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
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}