Question

How do you evaluate `\frac{1}{3}+2\log _{8}(3x-2)=\log _{8}(9x^{2}-24x+4)`?

Answer 1

No real solution.

Explanation:

`1/3+2log_8(3x-2)=log_8(9x^2-24x+4)` can be written as

`1/3log_8 8+2log_8(3x-2)=log_8(9x^2-24x+4)`

or `log_8 8^(1/3)+2log_8(3x-2)=log_8(9x^2-24x+4)`

or `log_8 2+log_8(3x-2)^2=log_8(9x^2-24x+4)`

or `2(3x-2)^2=9x^2-24x+4`

or `2(9x^2-12x+4)=9x^2-24x+4`

or `9x^2+4=0`

As for real values of `x`, `9x^2+4>0`, there is no real solution.