Answer
$n=a\left( e^{\frac{S}{a}}-1 \right)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
S=a\ln\left(1+\dfrac{n}{a}\right)
,$ for $
n
,$ use the properties of equality to isolate the $\ln$ expression. Then change to exponential form. Use the properties of equality to isolate the variable.
$\bf{\text{Solution Details:}}$
Dividing both sides by $a,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{S}{a}=\dfrac{a\ln\left(1+\dfrac{n}{a}\right)}{a}
\\\\
\dfrac{S}{a}=\ln\left(1+\dfrac{n}{a}\right)
.\end{array}
Since $\ln x =\log_e x,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{S}{a}=\log_e\left(1+\dfrac{n}{a}\right)
.\end{array}
Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to
\begin{array}{l}\require{cancel}
e^{\frac{S}{a}}=1+\dfrac{n}{a}
.\end{array}
Using the properties of equality, the equation above is equivalent to
\begin{array}{l}\require{cancel}
a\left(e^{\frac{S}{a}}\right)=a\left( 1+\dfrac{n}{a} \right)
\\\\
ae^{\frac{S}{a}}=a+n
\\\\
ae^{\frac{S}{a}}-a=n
\\\\
n=ae^{\frac{S}{a}}-a
\\\\
n=a\left( e^{\frac{S}{a}}-1 \right)
.\end{array}