How to solve $\ln(x) = 2x$ I know this question might be an easy one. but it has been so long since I solved such questions and I didn't find a an explanation on the internet. I'd like if someone can remind me. 
I reached that $e^{2x} = x$, but didn't know how to continue from here. I remember something that has to do with bases and equalizing parameters, but how do I do that in this case? 

Since $e^x\ge x+1$, we have
\begin{align}
e^{2x}&\ge(x+1)^2\\
&=x^2+2x+1\\
&=\left(x+\frac12\right)^2+\frac34+x\\
&>x
\end{align}
so there are no solutions (because they can never be equal).
I like using $e^x\ge x+1$ rather than calculus.
