# Question #99695

Jun 18, 2017

The logs more or less disappear!

#### Explanation:

${\left({x}^{2} + 6\right)}^{{\log}_{2} \left(x\right)} = {\left(5 x\right)}^{{\log}_{2} \left(x\right)}$

A special problem. If it were possible for the bases to be either positive or negative, we might encounter a case involving an absolute value. For now, don't worry about what that is. Why? Because the base "5x" is negative if and only if x is negative.

We know that x cannot be negative because ${\log}_{2} \left(x\right)$ is undefined unless $x > 0$. Also, ${x}^{2} + 6$ is always positive; therefore $5 x$ must be positive.

In this case...
${\left({x}^{2} + 6\right)}^{{\log}_{2} \left(x\right)} = {\left(5 x\right)}^{{\log}_{2} \left(x\right)}$
if and only if the bases are equal. That is,
${x}^{2} + 6 = 5 x$
Solve this as
${x}^{2} - 5 x + 6 = 0$
by using factoring.

Finish it off now. There are two solutions.