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…
5 Answers
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.
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 $x<0$.
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.
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.
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 $0.766$.
