Solving an equation in which the variable appears as both an exponent and a base I am currently playing around with exponential and logarithmic functions, and am now trying to solve the following: $2^x = x^2$.
My problem is that whenever I'm exponentiating or taking the logarithm of both sides, I realize that I'm going around in circles.
Is there any way on how you can isolate the $x$ for this equation, or any other equation in which the variable appears as both an exponent and a base?
Looking at the graphs, I can see that there are three intersections between the functions and the two rational solutions are $x=2$ and $x=4$ but I can't seem to be able to arrive at those answers algebraically.
 A: As percusse and GEdgar point-out in there comments that the reason this seemingly simple equation is not solvable using simple algebra lies in the fact that that the LHS of $2^x = x^2$ is a transcendental function. i.e. it cannot be expressed as a polynomial. Actually the closest it can come to in a "polynomial" form is its Maclaurin series form (see below).
Using pre-calculus techniques you can show, for instance, that you can take log of both sides as in
$$2^x = x^2$$
$$\implies ln(2^x) = ln(x^2) \quad \forall x \ne 0 $$
$$\implies x ln(2) = 2 ln(x) $$
$$\implies ln(x) = {2x \over ln(2)}  \quad \textbf {(A)}$$
So the solution to our problem are all values of $x$ that are the roots of equation $\textbf{(A)}$ ... Pre-calculus you can use graphing techniques to determine the answer.
Solving transcendental functions, in general, requires a lot of different calculus techniques, that are probably beyond the scope of this answer.
Infinite Series for ${2^x}$
Using Taylor's Theorem (which is part of calculus) we can show that:
$$e^u = \sum_{n=0}^{ \infty } {u^n \over n!} = 1 + {u^1 \over 1!} + {u^2 \over 2!} + {u^3 \over 3!} + {u^4 \over 4!} + \cdots  \quad \textbf{(B)}$$
For considerable historical reasons $\textbf{(B)}$ is called Maclaurin series for $e^u$. You can find Maclaurin series for a large number of functions that have certain properties.
For purpose of this discussion, assume that (B) is provable. We can use it to express the infinite series for $2^x$ by noting that $2 \equiv e^{ln(2)}$, and that $(a^x)^y = (a)^{xy}$.
$$ [2]^x = [e^{ln(2)}]^x = [e^{ln(2) \dot x}]$$
Substituting $u$ with $2^x$ in $\textbf{(B)}$ power-series we get:
$$2^x = ln(2) \sum_{n=0}^{ \infty } {x^n \over n!} = ln(2) \left( 1 + {x^1 \over 1!} + {x^2 \over 2!} + {x^3 \over 3!} + {x^4 \over 4!} + \cdots \right)$$
As you can see solving (A) without knowing some more properties about behavior of $e^x$ becomes intractable. That is what calculus is all about ;) once you get into it, you wil see that these problems become solvable.  Although, the solutions are, by no means, trivial.    
A: For your actual problem you can easily see that as well as $x=2$ and $x=4$, there is another solution in $[-1,0]$ and numerical methods will help you find it is about $x\approx -0.76666469596$.  Looking at the two curves $y=x^2$ and $y=2^x$, there are no other real solutions. That might be a good place to stop.
Alternatively, there is the Lambert W function or product logarithm with $z = W(z)\exp(W(z))$. With some manipulation, using $2^x=\exp(x\log(2))$ and $x^2=\exp(2\log(x))$ you can turn your original equation into $\left(- \frac{\log(2)}{2}x\right) \exp\left(- \frac{\log(2)}{2}x\right)   =  -
 \frac{\log(2)}{2}$ and so a solution is $- \frac{2}{\log(2)}W\left(-
 \frac{\log(2)}{2}\right)$, which Wolfram Alpha gives as 2.
But for negative real numbers $x^2=\exp(2\log(-x))$, and so with similar manipulation you should also look at solutions of the form  $- \frac{2}{\log(2)}W\left(
 \frac{\log(2)}{2}\right)$, which Wolfram Alpha gives as -0.76666...
To make life more complicated, the Lambert W function is multi-branched, so there can be more solutions, such as 4.  There are infinitely many complex solutions.
