# 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.

## 2 Answers

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 $(a^x)^y = (a)^{xy}$.

$$^x = [e^{ln(2)}]^x = [e^{ln(2) \dot x}]$$

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.

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.