egad...

logs don't work at all like that

what you could do is

$e^{2n} - 2e^n = 6$

$\ln\left(e^{2n} - 2e^n \right) = \ln(6)$

but this is as far as you can go using this strategy.

The actual way to solve this problem is with the substitution

$u = e^n$

Then we can rewrite the original equation as

$u^2 - 2u - 6 = 0$

Solve this for $u$

and then $x = \ln(u),~u>0$

If there are any solutions for $u$ such that $u\leq 0$ they can be ignored.