#### Answer

$\ln \frac{9}{10}$

#### Work Step by Step

$\Sigma^{\infty}_{n=1} e^{nc} =\Sigma^{\infty}_{n=1}(e^{c})^{n}$
$a_{1} = e^{0} =1$ and the common ratio is $r=e^{c}$
$\frac{1}{1-e^{c}}=10$
$1-e^{c} = \frac{1}{10}$
$-e^{c} = -\frac{9}{10}$
$e^{c} = \frac{9}{10}$
$c=\ln \frac{9}{10}$