Solve the following equation on the set of real numbers:

$$
\log _{19}(x-3)+\log _{93}(x-3)=3-\lg \left(x^{5}-24\right)
$$
The equation is meaningful if all three arguments of the logarithms are positive, that is, $x>3$ (in this case, $x^{5}-24>$ $3^{5}-24>0$ also holds).

On the half-line $(3, \infty)$, the left side is strictly monotonically increasing, as the base of each logarithm is greater than 1. On the right side, since $x^{5}-24$ and the base-10 logarithm are strictly monotonically increasing, the expression $3-\lg \left(x^{5}-24\right)$ is strictly monotonically decreasing. Since the value of a strictly monotonically increasing function and a strictly monotonically decreasing function can coincide at most at one point, the equation can have at most one solution.

It can be verified by substitution that this solution is $x=4$:

$$
\log _{19}(4-3)+\log _{93}(4-3)=\log _{19} 1+\log _{93} 1=0+0=0
$$

and

$$
3-\lg \left(4^{5}-24\right)=3-\lg 1000=3-3=0 .
$$