5. [5 points] Given the numbers $\log _{\sqrt{\frac{x}{3}+3}}(6 x-14), \log _{6 x-14}(x-1)^{2}, \log _{x-1}\left(\frac{x}{3}+3\right)$. For which $x$ are two of these numbers equal, and the third one less than them by 1?
Answer: $x=3$.

Solution. From the condition, it follows that the functions $6 x-14, x-1, \frac{x}{3}+3$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\sqrt{\frac{x}{3}+3}}(6 x-14), b=\log _{6 x-14}(x-1)^{2}, c=$ $\log _{x-1}\left(\frac{x}{3}+3\right)$. Then

$$
\begin{aligned}
& a b c=\log _{\sqrt{\frac{x}{3}+3}}(6 x-14) \cdot \log _{6 x-14}(x-1)^{2} \cdot \log _{x-1}\left(\frac{x}{3}+3\right)= \\
& \quad=2 \log _{\frac{x}{3}+3}(6 x-14) \cdot 2 \log _{6 x-14}(x-1) \cdot \frac{\log _{6 x-14}\left(\frac{x}{3}+3\right)}{\log _{6 x-14}(x-1)}=4
\end{aligned}
$$

By the condition, the numbers $(a ; b ; c)$ satisfy one of the three conditions:

$$
\begin{aligned}
& a=b \text { and } \quad a=c+1 \\
& b=c \text { and } c=a+1 \\
& c=a \text { and } \quad a=b+1
\end{aligned}
$$

Consider case (8). Substituting $b=a$ and $c=a-1$ into the equation $a b c=4$ obtained above, we have $a \cdot a \cdot(a-1)=4$, from which $a^{3}-a^{2}-4=0,(a-2)\left(a^{2}+a+2\right)=0$. Since the polynomial $a^{2}+a+2$ has no roots, the only solution to the equation is $a=2$, so the system is satisfied by the triplet of numbers $a=2, b=2, c=1$. Cases (9) and (10) are considered similarly; from them we get that either $a=1, b=2, c=2$, or $a=2, b=1, c=2$. Now, for each of the obtained triplets of numbers $(a ; b ; c)$, we will find $x$.

If $c=1$, then $x-1=\frac{x}{3}+3$, that is, $x=6$. Therefore, $a=2 \log _{\sqrt{5}} 22 \neq 2$, that is, there are no values of $x$ for which $a=b=2, c=1$.

If $a=1, b=c=2$, then $c=2 \Leftrightarrow(x-1)^{2}=\frac{x}{3}+3$. This equation is equivalent to the equation $3 x^{2}-7 x-6=0$, the roots of which are $x=3$ and $x=-\frac{2}{3}$, but $x=-\frac{2}{3}$ does not satisfy the domain of admissible values. For $x=3 a=2$, i.e., this root also does not fit in this case.

If $b=1$, then $(x-1)^{2}=(6 x-14)$. This equation is equivalent to the equation $x^{2}-8 x+15=0$, the roots of which are $x=3$ and $x=5$, but $x=5$ does not fit, as in this case $a=\log \sqrt{\frac{14}{3}} 16 \neq 2$. The value $x=3$ fits: $a=\log _{\sqrt{4}} 4=2, c=\log _{2} 4=2$.

Thus, $x=3$ is the only solution to the problem.