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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

Cristopher
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    note that there are two values, because $\ \sqrt(x)$ grows faster than 2logx – Mosk Sep 26 '14 at 18:13
  • If we are allowed to trust a calculator, proving there is, for example, a root between $2.04$ and $2.05$ is easy. We just show that $2\ln(x)-\sqrt{x}$ has different sign at $2.04$ than it does at $2.05$. (There is also a quite a bit larger root.) – André Nicolas Sep 26 '14 at 18:20

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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 $$x = e^{-2 W(-1/4)}.$$

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

Did
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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.

user_of_math
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