Answer
$a\left( e^{\frac {S}{a}}-1\right) $
Work Step by Step
$a\ln \left( 1+\dfrac {n}{a}\right) =S\Rightarrow \ln \left( 1+\dfrac {n}{a}\right) =\dfrac {S}{a}\Rightarrow 1+\dfrac {n}{a}=e^{\frac {S}{a}}\Rightarrow n=a\left( e^{\frac {S}{a}}-1\right) $
