2730.39 – Complicated Log Equation

There are many ways to solve this problem. One straight forward approach is to put all the terms of the equation on one side and combine them using the laws of logarithms.

Perhaps the simplest approach is to ask yourself: How can I get rid of the logarithms altogether?

The answer is that logs and exponential functions are inverse functions. If the base of the log and the base of the exponential are the same, then the logs disappear (exponentials too). Here is how to do this, one log at a time:

$$\begin{aligned}\log_4{2^{x^2}} &= \log_4{ \left( 4^{\left( \dfrac{1}{2}\right)} \right)^{x^2}} \\&= \log_4{ 4^{\left( \dfrac{x^2}{2} \right) }} \\&= \dfrac{x^2}{2}\end{aligned}$$

It works because $2$ is the square root of $4$. Similarly,

$$\log_4{2^{2x-1}} = x - \dfrac{1}{2}.$$

Thus the given equation becomes

$$\dfrac{x^2}{2} = 1 - \left(x - \dfrac{1}{2}\right).$$

This leads to the quadratic equation

$$x^2 + 2x - 3 = 0,$$

whose roots are 1 and -3. The answer is e.