Question #99695

1 Answer
Jun 18, 2017

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.