Solve the following equation:

$$
\frac{\log \left(35-x^{3}\right)^{3}}{\log (5-x)}=9
$$
The counter must have $\left(35-x^{3}\right)^{3}>0$, that is, $35-x^{3}>0, x1$, so $\log (5-x)>0$, which means the denominator and the quotient are defined. By multiplication:

$$
\log \left(35-x^{3}\right)^{3}=9 \log (5-x)
$$

From the equality of two (same base) logarithms, it follows that the numbers are also equal:

$$
\left(35-x^{3}\right)^{3}=(5-x)^{9}
$$

(In the realm of real numbers) the cube root is unique (a number has only one cube root), therefore

$$
35-x^{3}=(5-x)^{3}=125-75 x+15 x^{2}-x^{3} \text { and } x^{2}-5 x+6=0
$$

According to this, the roots of the equation can only be the numbers $x_{1}=2, x_{2}=3$. Both satisfy both the constraint and the equation.

Deák Margit (Körmend, Kölcsey F. g. I. o. t.)

Note: We can also arrive at (1) by considering the left side as the logarithm of the $\left(35-x^{3}\right)^{3}$ expression to the base $5-x$, and then taking into account the definition of the logarithm:

$$
\begin{aligned}
& 5-x \\
& \log \left(35-x^{3}\right)^{3}=9, \quad(5-x)^{9}=\left(35-x^{3}\right)^{3}
\end{aligned}
$$

Bóné András (Budapest, József A. g. IV. o. t.)