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How do I solve the equation $2^x - x^2 = 0$ for $x$ by using pen and paper? By rearranging the terms and taking $\log_2$ on both sides I obtain $x - \log_2 x^2 = 0.$

By using graphical calculator I know that there are three solutions. How do I proceed from the second equation to obtain these solutions? Maybe I am missing some trick here?

Doubloon Conroy
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    $\sqrt{2}$ is not a solution. There is no algebraic solution using the usual operations. – Paul Dec 27 '22 at 13:49
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    There is no direct way to calculate the solutions (if they cannot be guessed). We need numercial methods or the Lambert-W function might allow a closed form solution. – Peter Dec 27 '22 at 13:52
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    To emphasize, there are many questions which might be easy to ask but which are difficult to answer. Mixing exponentials and polynomials is exactly one such scenario. – JMoravitz Dec 27 '22 at 13:55
  • $\sqrt{2}$ is not a solution, that was my bad. – Doubloon Conroy Dec 27 '22 at 13:56
  • How about $x=2,4$? – Тyma Gaidash Dec 27 '22 at 14:03
  • The third solution (apart from $2$ and $4$) is $$-0.7666646959621230931112\cdots $$ – Peter Dec 27 '22 at 14:07
  • One way to solve for $x=2,4$ is:

    $x=\sqrt2,^x\implies x=\sqrt{2}^{\sqrt2^x}$ and iterate. Then use this result

     – Тyma Gaidash Dec 27 '22 at 14:16
  • The solutions are 2, 4 and -0.766... I naively assumed that there was a direct way to calculate the solutions. But apparantly one must use numerical methohds or as Peter, Ichoi and Abezhiko suggested by using Lambert W function. Thanks for the answers : ) – Doubloon Conroy Dec 27 '22 at 14:17

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With pen and paper, there is not much to do analytically apart from using the Lambert W function; it would go that way : $$ \begin{array}{lll} x^2 = 2^x = e^{x\ln 2} &\Longleftrightarrow& x^2e^{-x\ln 2} = 1 \\ &\Longleftrightarrow& xe^{-x\ln\sqrt{2}} = \pm 1 \\ &\Longleftrightarrow& (-x\ln\sqrt{2})e^{-x\ln\sqrt{2}} = \pm \ln\sqrt{2} \\ &\Longleftrightarrow& x_k^\pm = \displaystyle \frac{W_k(\pm\ln\sqrt{2})}{-\ln\sqrt{2}} \end{array} $$ where $k$ denotes the $k^{\mathrm{th}}$ branch of the Lambert W function. Otherwise, graphical methods are helpful, as mentioned by lChoi.

Abezhiko
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These equations aren't usually soluble with a standard method. I think for this equation, one must graph the two graphs $2^x$ and $x^2$ and bound the solutions first (as $x$ can't be that big, as $2^x$ grows much quicker than $x^2$), and then guess at them. We can then make sure we're not missing any solutions then by looking at the graphs again.

IChoi
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    Look at graphs "and then guess them"?? – amWhy Dec 27 '22 at 13:59
  • Yes - by looking at the graph you could ascertain how many solutions there are, and their rough size. Guessing and then verifying would be a valid strategy here. – IChoi Jan 30 '23 at 18:30