Problem 11.4. Find all values of the real parameter $a$ such that the equation

$$
\log _{a x}\left(3^{x}+4^{x}\right)=\log _{(a x)^{2}}\left(7^{2}\left(4^{x}-3^{x}\right)\right)+\log _{(a x)^{3}} 8^{x-1}
$$

has a solution.

Emil Kolev
11.4. Since $a x>0$ and $4^{x}-3^{x}>0$ it follows that $a>0$ and $x>0$.

For $a>0, x>0$ and $a x \neq 1$ the equation is equivalent to

$$
3^{x}+4^{x}=7.2^{x-1} \sqrt{4^{x}-3^{x}} \Longleftrightarrow 45\left(\frac{4}{3}\right)^{2 x}-57\left(\frac{4}{3}\right)^{x}-4=0
$$

Setting $y=\left(\frac{4}{3}\right)^{x}>0$ we obtain the equation $45 y^{2}-57 y-4=0$ with solutions $y_{1}=\frac{4}{3}$ and $y_{2}=-\frac{1}{15}$. Hence $x=1$. The condition $a x \neq 1$ now implies that $a \neq 1$

The required values of $a$ are $a \in(0,+\infty) \backslash\{1\}$.