
Problem 1. Find all positive roots of the equation

$$
\log _{(x+a-1)} \frac{4}{x+1}=\log _{a} 2
$$

where $a>1$ is a real number.


Solution. It is clear that if $a>1$ and $x>0$, then $x+a-1>0$ and $\frac{4}{x+1}>0$. Hence in this case $\log _{(x+a-1)} \frac{4}{x+1}$ is well defined. Since

$$
\log _{(x+a-1)} \frac{4}{x+1}=\frac{\log _{a} \frac{4}{x+1}}{\log _{a}(x+a-1)}
$$

then the given equation is equivalent to

$$
\frac{4}{x+1}=2^{\log _{a}(x+a-1)}
$$

The function $\frac{4}{x+1}$ is strictly decreasing in the interval $(0,+\infty)$. The function $\log _{a}(x+a-1)$ is strictly increasing in the interval $(0,+\infty)$ and obviously the same is true for the function $2^{\log _{a}(x+a-1)}$. Therefore the equation (1) has no more than one root in the interval $(0,+\infty)$.

On the other hand it is easy to check that $x=1$ is a root of this equation.
