slijmigrd

2022-07-12

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:

$x\mathrm{ln}\left(2\right)=2\mathrm{ln}\left(x\right)$

$\frac{\mathrm{ln}\left(2\right)}{2}}={\displaystyle \frac{\mathrm{ln}\left(x\right)}{x}$

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…

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:

$x\mathrm{ln}\left(2\right)=2\mathrm{ln}\left(x\right)$

$\frac{\mathrm{ln}\left(2\right)}{2}}={\displaystyle \frac{\mathrm{ln}\left(x\right)}{x}$

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…

SweallySnicles3

Beginner2022-07-13Added 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.

Joshua Foley

Beginner2022-07-14Added 3 answers

There is a special function, ${W}_{0}(x)$ that is the inverse of $f(x)=x{e}^{x}$ when the latter is restricted to $x\in [-1,\mathrm{\infty})$. Using this, expressions of the form $Y=X{e}^{X}$ can be solved as $X={W}_{0}(Y)$. You want to find the solution(s) to the equation ${2}^{x}={x}^{2}$. Rewrite ${2}^{x}$ as ${e}^{\mathrm{ln}(2)x}$ and raise each side to the power of $\frac{1}{2}$. We then arrive at

$x={e}^{\frac{\mathrm{ln}(2)}{2}x}$

Multiple both sides by $\frac{-\mathrm{ln}(2)}{2}{e}^{\frac{-\mathrm{ln}(2)}{2}}x$ to arrive at

$\frac{-\mathrm{ln}(2)}{2}x{e}^{\frac{-\mathrm{ln}(2)}{2}x}=\frac{-\mathrm{ln}(2)}{2}$

Apply ${W}_{0}$ to both sides to get

$\frac{-\mathrm{ln}(2)}{2}x={W}_{0}\left(\frac{-\mathrm{ln}(2)}{2}\right)$

Multiply through to find

$x=\frac{-2}{\mathrm{ln}(2)}{W}_{0}\left(\frac{-\mathrm{ln}(2)}{2}\right)$

Which is equal to 2.

$x={e}^{\frac{\mathrm{ln}(2)}{2}x}$

Multiple both sides by $\frac{-\mathrm{ln}(2)}{2}{e}^{\frac{-\mathrm{ln}(2)}{2}}x$ to arrive at

$\frac{-\mathrm{ln}(2)}{2}x{e}^{\frac{-\mathrm{ln}(2)}{2}x}=\frac{-\mathrm{ln}(2)}{2}$

Apply ${W}_{0}$ to both sides to get

$\frac{-\mathrm{ln}(2)}{2}x={W}_{0}\left(\frac{-\mathrm{ln}(2)}{2}\right)$

Multiply through to find

$x=\frac{-2}{\mathrm{ln}(2)}{W}_{0}\left(\frac{-\mathrm{ln}(2)}{2}\right)$

Which is equal to 2.

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