Problem 1. Determine the real numbers $x \in(2, \infty)$, which are solutions to the equation

$$
\cos \left(\pi \log _{3}(x+6)\right) \cdot \cos \left(\pi \log _{3}(x-2)\right)=1
$$

Supplement to Gazeta Matematică
Solution: The hypothesis leads to $\cos \left(\pi \log _{3}(x+6)\right)=\cos \left(\pi \log _{3}(x-2)\right)= \pm 1$, so there exist $k, l \in \mathbb{Z}$, of the same parity, such that $\pi \log _{3}(x+6)=k \pi$ and $\pi \log _{3}(x-2)=l \pi$. .. 2p

We obtain the relations $x+6=3^{k}$ and $x-2=3^{l}$. By subtraction, we are led to

![](https://cdn.mathpix.com/cropped/2024_06_07_b5f4bd8b782e050f8d06g-2.jpg?height=43&width=1207&top_left_y=1537&top_left_x=459)

The equality is not possible if $k, l<0$. Then $3^{l}\left(3^{k-l}-1\right)=8$, from which $3^{l}=1$, i.e., $l=0$ and then $k=2$. Finally, we obtain $x=3$. ................ $\mathbf{p}$