## PROBLEM 2

Solve the equation

$$
\log _{4}\left(5^{x}-1\right)=\log _{5}\left(4^{x}+1\right)
$$
Solution: Condition: $5^{x}-1>0 \Rightarrow x \in(0, \infty)$

Let $\quad \log _{4}\left(5^{x}-1\right)=t \Rightarrow 5^{x}-1=4^{t}$.

$\log _{5}\left(4^{x}+1\right)=t \Rightarrow 4^{x}+1=5^{t}$.

Adding the two relations above results in $4^{x}+5^{x}=4^{t}+5^{t}$ $1 p$

Since the function $f: \square \rightarrow(0, \infty), f(x)=4^{x}+5^{x}$ is strictly increasing, hence injective, it follows that $x=t$.

Then $\quad 5^{x}-1=4^{x}$, from which $\left(\frac{4}{5}\right)^{x}+\left(\frac{1}{5}\right)^{x}=1$. (1) $1 p$

The function $g: \square \rightarrow(0, \infty), \quad g(x)=\left(\frac{4}{5}\right)^{x}+\left(\frac{1}{5}\right)^{x}$ is strictly decreasing, therefore $g$ is injective, so equation (1) has at most one solution, which results in $x=1$ being the unique solution.