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I have been thinking about this equation:

$$x^2=2^x$$

I know there is two integer solutions: $x=2$ and $x=4$. But there also is a negative solution, that is approximately $x=-0.77$.

$$(-0.77)^2=0.5929$$ $$2^{(-0.77)}=0.5864...$$

Can we find this negative solution exactly?

Roberto Gandulfo
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2 Answers2

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The solutions for the equation $x^{a}=a^{x}$ are given by $x=-\frac{a}{\log a}W\left(-\frac{\log a}{a}\right)$, where $W(\cdot)$ is the Lambert W-function, i.e., the inverse function of $f(W)=We^{W}$. Then we get $x_{1}=2,x_{2}=4, x_{3}\approx-0.7\bar{6}$.

P.S: To obtain the value $x_{3}$ we need to consider $x<0$, we get $x=-\frac{a}{\log a}W\left( \frac{\log a}{a}\right)$ in our case $a=2$ give $x_{3}=-\frac{2}{\log 2}W\left( \frac{\log 2}{2}\right)\approx -0.7\bar{6}$.

A. P.
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  • How do you get $x_3$ from $-\frac{2}{\log 2}W\left(-\frac{\log 2}{2}\right)$? I thought $W(x)$ had two values, both negative, when $-e^{-1}<x<0$, and when you multiply these by a negative number such as $-\frac{2}{\log 2}$ you get positive results as in $2$ or $4$ here – Henry Feb 19 '22 at 00:55
  • @Henry You are right and thank you for the remark. I was reading your comment and I've been working on $x_{3}$ and I hope it's clear now. I corrected. – A. P. Feb 19 '22 at 13:33
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    Indeed (+1). The negative solution is then close to $-0.766664695962123\ldots$ – Henry Feb 19 '22 at 14:53
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$x$ is a solution of the equation $x^2 = 2^x$ if : $$\dfrac{\ln|x|}{x} = \dfrac{\ln 2}{2}$$ from the graph of the function : $$f(x) = \dfrac{\ln|x|}{x} - \dfrac{\ln 2}{2}$$ enter image description here

we deduce that there are only $3$ solutions $x = 2$, $x = 4$ and $x = -0.767$.

Essaidi
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