I need help solving $3e^{2x}-1=\frac{1}{2}$

Question

I am trying to solve $3e^{2x}-1=\frac{1}{2}$. Here is my work:

$3e^{2x}-1=\frac{1}{2}$

$3e^{2x} =\frac{1}{2}+1$

$e^{2x} =\frac{1.5}{3}$

$\ln{e^{2x}} =\ln{(\frac{1}{6})}$

$2x =\ln(\frac{1}{6})$

$x =\frac{\ln(\frac{1}{6})}{2}$

$x =-0.895$

When I check my answer I get $-\frac{1}{2}$.

Please show what I am doing wrong.

Answer 1

$\frac{1.5}{3}=\frac{1}{2}$, not $\frac{1}{6}$.