$(\log_23) ^ x + (\log_35) ^ x = 2 (\log_34) ^ x$

Question

So, I have encountered this equation and I do not know how to solve it. $$(\log_23) ^ x + (\log_35) ^ x = 2 (\log_34) ^ x$$ My first idea was dividing the equation by $\log_34$ and getting into something like $\left(\frac{\log_23}{\log_34}\right)^ x + \left(\frac{\log_35}{\log_34}\right) ^ x = 2$. The obvious solution is $x = 0$ and I tried proving that LHS is smaller than $1$ if $x \lt 1 $ and bigger if $x > 1$, but realized $\frac{\log_23}{\log_34} \lt 1$, so my proof does n

Answer 1

$$\left(\frac{\log _3 5}{\log _3 4}\right)^x+\left(\frac{\log _2 3}{\log _3 4}\right)^x=2$$ has only the root $x=0$.

Define the function $$f(x)=\left(\frac{\log _3 5}{\log _3 4}\right)^x+\left(\frac{\log _2 3}{\log _3 4}\right)^x$$ $f(x)$ is always increasing because, approximating to simplify the writing, we have $$f(x)\approx 1.16^x+1.26^x\to f'(x)\approx 0.15\cdot 1.16^x+0.23\cdot1.26^x>0, \forall x\in\mathbb{R}$$ Therefore before and after the root $x=0$ there are no other roots.