# Solve 2 x </msup> = x 2 </msup> I've been asked to solve th

slijmigrd 2022-07-12 Answered
Solve ${2}^{x}={x}^{2}$
I've been asked to solve this and I've tried a few things but I have trouble eliminating $x$. I first tried taking the natural log:
$x\mathrm{ln}\left(2\right)=2\mathrm{ln}\left(x\right)$
$\frac{\mathrm{ln}\left(2\right)}{2}=\frac{\mathrm{ln}\left(x\right)}{x}$
I don't know what to do from here so I decided to try another method:
${2}^{x}={2}^{{\mathrm{log}}_{2}\left({x}^{2}\right)}$
$x={\mathrm{log}}_{2}\left({x}^{2}\right)$
And then I get stuck here, I'm all out of ideas. My guess is I've overlooked something simple…
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## Answers (2)

SweallySnicles3
Answered 2022-07-13 Author has 21 answers
Your equation has two obvious solutions which are $x=2$ and $x=4$. The last solution is not rational ($x\approx -0.766665$) and cannot be obtained using simple functions. You cannot get the last root using logarithms.
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Joshua Foley
Answered 2022-07-14 Author has 3 answers
There is a special function, ${W}_{0}\left(x\right)$ that is the inverse of $f\left(x\right)=x{e}^{x}$ when the latter is restricted to $x\in \left[-1,\mathrm{\infty }\right)$. Using this, expressions of the form $Y=X{e}^{X}$ can be solved as $X={W}_{0}\left(Y\right)$. You want to find the solution(s) to the equation ${2}^{x}={x}^{2}$. Rewrite ${2}^{x}$ as ${e}^{\mathrm{ln}\left(2\right)x}$ and raise each side to the power of $\frac{1}{2}$. We then arrive at
$x={e}^{\frac{\mathrm{ln}\left(2\right)}{2}x}$
Multiple both sides by $\frac{-\mathrm{ln}\left(2\right)}{2}{e}^{\frac{-\mathrm{ln}\left(2\right)}{2}}x$ to arrive at
$\frac{-\mathrm{ln}\left(2\right)}{2}x{e}^{\frac{-\mathrm{ln}\left(2\right)}{2}x}=\frac{-\mathrm{ln}\left(2\right)}{2}$
Apply ${W}_{0}$ to both sides to get
$\frac{-\mathrm{ln}\left(2\right)}{2}x={W}_{0}\left(\frac{-\mathrm{ln}\left(2\right)}{2}\right)$
Multiply through to find
$x=\frac{-2}{\mathrm{ln}\left(2\right)}{W}_{0}\left(\frac{-\mathrm{ln}\left(2\right)}{2}\right)$
Which is equal to 2.
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