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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: $$x\ln \left( 2\right) =2\ln \left( x\right)$$ $$\dfrac {\ln \left( 2\right) }{2}=\dfrac {\ln \left( x\right) }{x}$$ I don't know what to do from here so I decided to try another method: $$2^{x}=2^{\log _{2}\left( x^{2}\right) }$$ $$x=\log _{2}\left( x^{2}\right)$$ And then I get stuck here, I'm all out of ideas. My guess is I've overlooked something simple…

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    $\begingroup$ Can you guess some solutions? You are not going to be able to "solve" the equation explicitly with precalculus tools. $\endgroup$ – Andrés E. Caicedo Dec 3 '13 at 16:07
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    $\begingroup$ If you can graph $\ln(x)/x$, you will see that the equation has exactly two solutions. (Using calculus, this can be proved formally.) $\endgroup$ – Andrés E. Caicedo Dec 3 '13 at 16:08
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    $\begingroup$ @Andres Caicedo the problem with $\ln(x)/x$ = $\ln(2)/2$ is that it doesn't accept negative $x$ as solutions, whereas there does exist a nontrivial negative solution to $2^x = x^2$, in addition to the (relatively) obvious positive integer roots. $\endgroup$ – Zubin Mukerjee Dec 3 '13 at 16:14
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    $\begingroup$ en.wikipedia.org/wiki/Lambert_W_function $\endgroup$ – Alex Dec 3 '13 at 16:17
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    $\begingroup$ @ZubinMukerjee Ah, yes, good point! Thanks. Besides $\ln(2)/2$, one may want to consider also $-\ln(2)/2$, because $\ln (x)/x =-\ln(2)/2$ is equivalent to $2^{-x}=(-x)^2$. $\endgroup$ – Andrés E. Caicedo Dec 3 '13 at 16:25
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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.

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    $\begingroup$ More precisely -0.76666469596212.. $\endgroup$ – Tomas Dec 3 '13 at 18:38
  • $\begingroup$ The only function that gives it as an exact answer is a trancendental one, the lambert omega function, given in one of the answers below. $\endgroup$ – Alan Sep 22 '16 at 15:44
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Consider the function$$f(x):=(\ln 2)x-2\ln x$$ then $f^\prime (x)=\ln 2-2/x$. Then it easily follows that $f^\prime (x)>0$ when $x>4$ and $f^\prime (x)< 0$ when $x<2$. That is $f$ is increasing when $x>4$ and it is decreasing when $x<2$. Also $4$ and $2$ are zeros of $f$. Hence it follows that these are the only zero for $x>0$.

For, $x<0$ put $x=-y$ and consider the function $$g(y)=-(\ln 2)y-2\ln y$$ Then $g^\prime (y)=-\ln 2-2/y<0$ for all $y>0$ i.e. the function is strictly decreasing and hence it has exactly one root for $x<0$.

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    $\begingroup$ Going to logarithms, you miss one root. $\endgroup$ – Claude Leibovici Dec 3 '13 at 16:21
  • $\begingroup$ @Claude Leibovici: Thanks for pointing that out, I have edited the answer. $\endgroup$ – pritam Dec 3 '13 at 16:32
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There is a special function, $W_0(x)$ that is the inverse of $f(x)=xe^x$ when the latter is restricted to $x\in [-1,\infty)$. Using this, expressions of the form $Y=Xe^X$ 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 $e^{\ln(2)x}$ and raise each side to the power of $\frac{1}{2}$. We then arrive at $$x=e^{\frac{\ln(2)}{2}x}$$Multiple both sides by $\frac{-\ln(2)}{2}e^{\frac{-\ln(2)}{2}}x$ to arrive at $$\frac{-\ln(2)}{2} x e^{\frac{-\ln(2)}{2}x}=\frac{-\ln(2)}{2}$$Apply $W_0$ to both sides to get $$\frac{-\ln(2)}{2}x=W_0\left(\frac{-\ln(2)}{2}\right)$$ Multiply through to find $$x=\frac{-2}{\ln(2)} W_0\left(\frac{-\ln(2)}{2}\right)$$Which is equal to 2.

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  • $\begingroup$ As a side note, the name of this function (Family of functions) is the Lambert Omega function, if you want to search for it. This problem plagued me for almost 2 decades before I learned the solution :) $\endgroup$ – Alan Sep 22 '16 at 15:42
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Resolution graphics:$$x^2=2^x<=>|x|=2^{\frac{x}{2}}$$

In the interval $ (-\infty,0) $ the equation has a solution because the member function is strictly decreasing and the left from the right hand is strictly increasing. In the interval $(0, \infty)$, the equation has two solutions $2$ and $4$, the function of the left hand side is linear function and the function of the right hand is convex.

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It is interesting to see that the two functions actually have three intersections. I would add that the non rational solution can be approximated by an iterative representation. Here I use $2^{-x} = x^2$, the solution to this equation is just negative of the solution to the original problem. The iteration would be: $$ x_{n+1} = \frac{1}{2^{x_n/2}} $$ Start with $x_0 = 1$ and five iterations give you $0.766$.

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