## Task B-3.1.

Determine all integers $x$ such that $\log _{3} \frac{2 x+60}{x+5}$ is also an integer.
## Solution.

From $\frac{2 x+60}{x+5}>0$ we get $x \in\langle-\infty,-30\rangle \cup\langle-5,+\infty\rangle$.

We have three possibilities:

$\log _{3} \frac{2 x+60}{x+5}=0$ or $\log _{3} \frac{2 x+60}{x+5}=n$ or $\log _{3} \frac{2 x+60}{x+5}=-n, n \in \mathbb{N}$.

Let's check if there is an integer $x$ such that $\log _{3} \frac{2 x+60}{x+5}=0$.

$\frac{2 x+60}{x+5}=1 \Rightarrow x=-55$.

Let $\log _{3} \frac{2 x+60}{x+5}=n$, where $n$ is a natural number. Then $\frac{2 x+60}{x+5}=3^{n}$.

$\frac{2 x+60}{x+5}=\frac{2(x+5)+50}{x+5}=2+\frac{50}{x+5} \Rightarrow \frac{50}{x+5}=3^{n}-2$.

Since $3^{n}-2$ is an odd number, due to divisibility, we have $x+5 \in\{2,10,50\}$, or $x \in$ $\{-3,5,45\}$.

By substitution, it is verified that the solutions are $x=-3, x=45$.

Now let $\log _{3} \frac{2 x+60}{x+5}=-n$, where $n$ is a natural number. Then

$$
\begin{gathered}
\frac{2 x+60}{x+5}=3^{-n} \Leftrightarrow \frac{x+5}{2 x+60}=3^{n} \Leftrightarrow \frac{x+5}{x+30}=2 \cdot 3^{n} \\
\Leftrightarrow \frac{x+30-25}{x+30}=2 \cdot 3^{n} \Leftrightarrow 1-\frac{25}{x+30}=2 \cdot 3^{n}
\end{gathered}
$$

Since the right side is an even integer, on the left side we must have $x+30 \in\{-1,-5,-25\}$, or $x \in\{-31,-35,-55\}$.

By substitution, it is verified that the solutions are $x=-35$ and $x=-55$.

Therefore, the sought integers $x$ are $-55, -35, -3$, and 45.