# 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).$$[x^{-2}e^{\ln(2)x}]^{-\frac12}=^{-\frac12}$$$$xe^{-\frac12\ln(2)x}=1\to-\frac12\ln(2)xe^{-\frac12\ln(2)x}=-\frac12\ln(2)$$

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.

The equation you presented is what is known as a 'transcendental' equation - the variable you're trying to isolate exists in two different classes of functions. On the left hand side it is in an algebraic function, while on the right hand side it is an exponential. In general when this happens, it is impossible to isolate $x$ on one side of the equation. Like you indicated, the only option to obtain numerical solutions would be graphically or using an iterative algorithm such as the bisection method or the secant method.

It is actually possible to obtain approximate closed form analytic solutions for such equations using the methods of calculus. It sounds like this would be a little bit out of your depth at the moment, but if you're interested and when you feel you're ready, look up asymptotic analysis or perturbation methods.

• Wow, sounds pretty intense. Thanks for the answer. I'll do some more research. Thanks. – Anson Savage Dec 12 '15 at 2:25
• No problem - stay curious! – pr0gramR Dec 12 '15 at 2:33
• I'll look those up. I always deal with exponentials with the Lambert W function though. – Simply Beautiful Art Dec 14 '15 at 1:26
• The W function is interesting and useful, but the questioner was clearly approaching the problem from a precalculus point of view. So I assume what he was really looking for was a method to arrive at numerical solutions using only elementary algebraic manipulations, which the Lambert function circumvents (and is thus nonelementary). – pr0gramR Dec 14 '15 at 2:53

Just for your curiosity, there are two trivial solutions for $x^2=2^x$, namely $x_1=2$ and $x_2=4$. Concerning the negative root, it is given by $$x_3=-\frac{2 }{\log (2)}\,W\left(\frac{\log (2)}{2}\right)\approx -0.766665$$ where appears Lambert function. You will (sooner or later) learn about this function which has a lot of practical applications.

In fact, any equation which can write $A+Bx+C\log(D+Ex)=0$ has solutions (real and/or complex) which can be expressed in terms of Lambert function.