$e$ and natural logarithms

Question

How would you solve $6xe^{2x}+3e^{2x}=0$ for $x$

I tried:

$\ln(e^{2x})=\ln(1/6x+3)$

$2x=\ln(1)-\ln(6x+3)$

$2x=-\ln(6x+3)$

but then I am stuck there.

What am I missing?

Answer 1

$$3e^{2x}(2x+1)=0 \Rightarrow x=-\frac{1}{2}, \text{ as } e^{2x} \neq 0, \forall x$$