$\displaystyle cosh(x) = \frac{5}{3} $, solve for x.

This is how I go about it (not sure if it's the right start).

$\displaystyle \frac{1}{2}(e^x + e^{-x}) = \frac{5}{3} $

$\displaystyle e^x + e^{-x} = \frac{10}{3} $

Is this next step right?

$\displaystyle x + \frac{1}{x} = log_e10 - log_e3 $

If so, how can I solve for x from here? Cheers,