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What are the roots of $ x^{2} = 2^{x} $? I drew the graphs and found $ x = 2 $ and $ x = 4 $, and there is one other root in $ [-1,0] $. Can anyone describe an algebraic method to obtain all roots?


marked as duplicate by user147263, Mark Viola, Claude Leibovici calculus Dec 1 '15 at 7:07

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    $\begingroup$ This question was already asked here. There is no nice formula to get the solution. You either need numerical methods or the Lambert-W-Function. $\endgroup$ – Peter Nov 30 '15 at 19:03
  • $\begingroup$ The negative solution is $-0.7666646959621230931112044225$ $\endgroup$ – Peter Nov 30 '15 at 19:05

Taking the logarithm and assuming $x>0$,

$$2\ln(x)=x\ln(2),$$ or $$\frac{\ln(x)}x=\frac{\ln(2)}2.$$

The derivative of the LHS is $$\frac{1-\ln(x)}{x^2},$$ which has a single root at $x=e$.

As there is a single extremum and the function is continuous, the equation

$$\frac{\ln(x)}x$$has at most two roots, which you found.

For negative $x$, the equation turns to


As the LHS function is positive only on one side of the extremum, there is at most one negative root. This root exists as the function goes to $\infty$, but it has no closed form.

  • $\begingroup$ With this method, we do not get the negative solution. $\endgroup$ – Peter Nov 30 '15 at 19:07
  • $\begingroup$ It has at most two positive solutions. But if $x<0$ then $\ln(x^2)$ is defined and $\ln(x)$ is not, so $\ln(x^2)=2 \ln(x)$ fails there. $\endgroup$ – Ian Nov 30 '15 at 19:07
  • $\begingroup$ That's right, should be $2\ln(|x|)$. $\endgroup$ – Yves Daoust Nov 30 '15 at 19:15

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