How to solve $\ln(x) = 2x$

Question

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?

Answer 1

Draw a graph. $\log x < 2x $

A proof is by noting that $\log x < 2x$ for $x < 1$ and then differentiating both sides to see that the LHS grows slower than the RHS.

Equivalently, $e^{2x} > x$

Answer 2

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.

Answer 3

This equation has no solution. To see it, note that $\ln(x)<x$$<2x$ for every $x>0$.

Answer 4

Assuming you want to solve this for complex $x$:

The function $f(z) = ln(z)-2z$ has an essential singularity at 0 and by Casorati-Weierstrass takes every complex value infinitely many times. And therefore must have infinitely many complex solutions there.