Answer
$x=4$ and $x=2$
Work Step by Step
$2\log x=\log2+\log(3x-4)$
Rewrite the left side of the equation by raising the argument of the $\log$ to the power of $2$:
$\log x^{2}=\log2+\log(3x-4)$
Now, combine the logarithms on the right side as a product:
$\log x^{2}=\log2(3x-4)$
$\log x^{2}=\log(6x-8)$
Since $\log$ is one to one, this equation becomes:
$x^{2}=6x-8$
Take all terms to the left side:
$x^{2}-6x+8=0$
Solve by factoring:
$(x-4)(x-2)=0$
We get two solutions:
$x=4$ and $x=2$
