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This question already has an answer here:

So lately I came across this seemingly simple problem that I just can't get around. Solve this equation:

$x^2 = 2^x$

I cannot do this algebraically, while I refuse to believe it is impossible to solve this algebraically or at least without a graphic calculator (maybe use Newton's method or series approximations?).

I decided to simplify the problem:

$x = 2^x$

Again, I cannot find an exact way to solve this.

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marked as duplicate by Aaron Maroja, Ali Caglayan, Lord_Farin, N. F. Taussig, dustin Feb 6 '15 at 0:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ you can only solve this equation asymptotically for $x\rightarrow 0$ or $x\rightarrow \infty$.Or just guess that x=2 :) $\endgroup$ – tired Feb 5 '15 at 18:39
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    $\begingroup$ Note that $x=4$ is also a solution. $\endgroup$ – Tim Raczkowski Feb 5 '15 at 18:46
  • $\begingroup$ What do you mean solve asymptotically? $\endgroup$ – user209347 Feb 5 '15 at 18:47
  • $\begingroup$ Check these answers. $\endgroup$ – Aaron Maroja Feb 5 '15 at 20:05
  • $\begingroup$ The same question is treated here: math.stackexchange.com/questions/591124/solve-2x-x2 $\endgroup$ – Chris Culter Mar 23 '16 at 23:17
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If you rearrange both sides and take logarithms:

$$e^{\ln(x)\cdot2} = 2^x$$

$$2\cdot\ln(x)=x\cdot\ln(2)$$

This type of equation, a relationship between a polynomial and a logarithm has been shown to have no solutions in standard analytic functions. As such, a new function - the Lambert $W$ function - was defined as the $W(z)=y$ such that $y=z\cdot e^z$.

In this case the problem can be solved by seeing that the equations have a symmetry in $x$ and $2$, hence $x=2$ is a solution.

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Hint: You have a "2" in one of the exponents and a "2" in one of the bases, and the rest of the symbols are "x". So what should you plug in for $x$ to make all the symbols equal?

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  • $\begingroup$ Okay, but that is not really algebra right?? I mean, it definitely is a solution, but not all the solutions. Its the same as saying $x^2 = 2^2$ in order to make all equal, so x = 2 is the solution. But we all know x = -2 is another solution. $\endgroup$ – user209347 Feb 5 '15 at 18:43
  • $\begingroup$ Yeah of course, Im referring to the example in the comment, not the actual question. $\endgroup$ – user209347 Feb 5 '15 at 18:46
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There are two trivial solutions : $x=2$ and $x=4$

The non trivial solution can be computed with numerical methods or analyticaly, thanks to the Lambert W function : $$x^2=2^x=e^{x\ln(2)}$$ $$x e^{-\frac{\ln(2)}{2}x}=1$$ with $x=-\frac{2}{\ln(2)}X$ $$Xe^X=-\frac{\ln(2)}{2}$$ $$X=W\left(-\frac{\ln(2)}{2}\right)$$ $$x=-\frac{2}{\ln(2)}W\left(-\frac{\ln(2)}{2}\right)\simeq 0.766665$$

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