How to solve $2\ln(x) = \sqrt{x}$ ? ln = natural logarithm

Question

I used Microsoft Mathematics and it says $x$ is approximately $2.04\dots$ but, how do you prove it?

Edit: I'm sorry if I wasn't clear enough with my question. I don't want to prove that two roots exist. What I would like is a way to find the approximate value of these two roots. Thanks to everyone who answered

Answer 1

To expand on @Did's answer: the Lambert $W$-function is the inverse of $y = xe^x$, $x \ge -1$. Thus $x = W(y)$ if and only if $y \ge -\frac 1e$ and $y = xe^x$.

Let $2 \ln x = \sqrt x$ and let $t = \ln x$. Then $2t = \sqrt{e^t} = e^{t/2}$, so $-\frac 12 te^{-t/2} = -\frac 14$. There are two solutions to this equation, but the solution $t$ with $-\frac t2 > -1$ satisfies $-\frac t2 = W(-\frac 14)$ so that $t = -2 W(-\frac 14)$. Thus one solution is $$x = e^{-2 W(-1/4)}.$$

Answer 2

Probably $x=16\cdot W^2(-1/4)$...

Answer 3

For the second part of your question, you can prove that a solution exists, and also bracket the root using calculus.

For x=1, 2 log(x) = 0 < $\sqrt{x}$ = 1

For x=e, 2 log(e) = 2 > $\sqrt{e}$ since $e < 4$ and sqrt(x) is a monotonically increasing function.

So by the intermediate value theorem $2 \log(x) - \sqrt{x}$ has a root between 1 and e.

You can, of course, tighten the bracket around the root further.