2730.39 – Complicated Log Equation

For what values of xx is the following statement true?

log⁡42x2=1−log⁡422x−1.\log_4{2^{x^2}} = 1 - \log_4{2^{2x-1}}.

  1. -2, 4
  2. 2, -4
  3. -1, -3
  4. 1, 3
  5. 1, -3
Solution

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:

log⁡42x2=log⁡4(412)x2=log⁡44(x22)=x22 \begin{aligned} \log_4{2^{x^2}} &=& \log_4{ \left( 4^{\frac{1}{2}} \right)^{x^2}} \\ &=& \log_4{ 4^{\left( \frac{x^2}{2} \right) }} \\ &=& \frac{x^2}{2} \end{aligned}

It works because 22 is the square root of 44. Similarly,

log⁡422x−1=x−12.\log_4{2^{2x-1}} = x - \frac{1}{2}.

Thus the given equation becomes

x22=1−(x−12).\frac{x^2}{2} = 1 - \left(x - \frac{1}{2}\right).

This leads to the quadratic equation

x2+2x−3=0,x^2 + 2x - 3 = 0,

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