How to solve a problem with a variable in both the base and exponent on opposite sides of an equation I am working on systems of equations in Pre-Calculus, and I presented the teacher a question that I had been wondering for a while.
$x^2 = 2^x$
The teacher couldn't figure it out after playing with it for quite a while.  What are some ways it can be solved algebraically?  Of course it can be solved by graphing, but what about for exact answers or possibly imaginary solutions?  If the answer could include a step by step solution, that would be great.  Thanks for the help.

Love the curiosity!
To solve this requires more than regular algebra.  It requires use of the Lambert W function.$$f(x)=xe^x$$$$W(x)=f^{-1}(x)$$The solution, is, of course, not solvable.  But it does allow us to do some amazing things.
First, let's attempt to solve for $W(x)$ to find its identities.$$x=ye^y$$$$y=W(x)$$Upon using substitutions, we get two identities.$$(1)y=W(ye^y)$$$$(2)x=W(x)e^{W(x)}\to\frac x{W(x)}=e^{W(x)}$$
Now, lets try to solve.$$2^x=x^2$$First, note that we must have base $e$.$$e^{\ln(2)x}=x^2\to\frac{e^{\ln(2)x}}{x^2}=1\to x^{-2}e^{\ln(2)x}=1$$Now the whole point is to get the base and exponent to be the same so that we can use the first identity(1).$$[x^{-2}e^{\ln(2)x}]^{-\frac12}=[1]^{-\frac12}$$$$xe^{-\frac12\ln(2)x}=1\to-\frac12\ln(2)xe^{-\frac12\ln(2)x}=-\frac12\ln(2)$$
Now we take the "$W$" of both sides to produce the first identity(1).$$W(-\frac12\ln(2)xe^{-\frac12\ln(2)x})=W(-\frac12\ln(2))$$$$-\frac12\ln(2)x=W(-\frac12\ln(2))$$
Now divide.$$x=\frac{W(-\frac12\ln(2))}{-\frac12\ln(2)}$$
Use a calculator to find all values.
Also, this allows an infinite number of complex answers.
