Equation $\log_\sqrt{3}(\log_2(x))+2\log_\sqrt{3}(\log_4(3x-2))=0$

Question

$\log_\sqrt{3}(\log_2(x))+2\log_\sqrt{3}(\log_4(3x-2))=0$

I was able to bring to $\log_2(x)\log_4^2(3x-2))=1$, but I don't know what to do next.

Thanks.

Answer 1

The function $f: \left( \frac{3}{2} , \infty \right) \to \mathbb{R}_+$, defined by $f(x)=\log_2 (x) \cdot \log_4^2 (3x-2)$ is (strictly) increasing (because it is a product of 2 (strictly) increasing functions). So, the equation $f(x)=1$ has at most one solution in $\left( \frac{3}{2} , \infty \right) $, which is $x=2$.

P.S.: f is defined on $\left( \frac{3}{2} , \infty \right) $ because we must have $x>0$ and $3x-2>0$ (for the existence of logarithms).