Question

Question #99695

Answer 1

The logs more or less disappear!

Explanation:

`(x^2 + 6)^(log_2(x)) = (5x)^(log_2(x))`

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(x)` is undefined unless `x > 0`. Also, `x^2 + 6` is always positive; therefore `5x` must be positive.

In this case...
`(x^2 + 6)^(log_2(x)) = (5x)^(log_2(x))`
if and only if the bases are equal. That is,
`x^2 + 6 = 5x`
Solve this as
`x^2 -5x + 6 = 0`
by using factoring.

Finish it off now. There are two solutions.