5. [5 points] Given the numbers $\log _{\sqrt{2 x-8}}(x-4), \log _{(x-4)^{2}}(5 x-26), \log _{\sqrt{5 x-26}}(2 x-8)$. For which $x$ are two of these numbers equal, and the third one greater than them by 1?
Answer: $x=6$.

Solution. From the condition, it follows that the functions $x-4, 5x-26$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\sqrt{2 x-8}}(x-4), b=\log _{\sqrt{5 x-26}}(2 x-8), c=\log _{(x-4)^{2}}(5 x-26)$. Then

$$
\begin{aligned}
& a b c=\log _{\sqrt{2 x-8}}(x-4) \cdot \log _{\sqrt{5 x-26}}(2 x-8) \cdot \log _{(x-4)^{2}}(5 x-26)= \\
& \quad=2 \log _{2 x-8}(x-4) \cdot 2 \log _{5 x-26}(2 x-8) \cdot \frac{\log _{2 x-8}(5 x-26)}{\log _{2 x-8}(x-4)^{2}}=2
\end{aligned}
$$

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 (18). Substituting $b=a$ and $c=a+1$ into the equation $a b c=2$, we have $a \cdot a \cdot(a+1)=2$, from which $a^{3}+a^{2}-2=0,(a-1)\left(a^{2}+2 a+2\right)=0$. Since the polynomial $a^{2}+2 a+2$ has no roots, the only solution to the equation is $a=1$, so the system is satisfied by the triplet of numbers $a=1, b=1, c=2$. Cases (19) and (20) are considered similarly; from them we get that either $a=2, b=1, c=1$, or $a=1, b=2, c=1$. Now, for each of the obtained triplets of numbers $(a ; b ; c)$, we will find $x$.

If $b=2$, then $5 x-26=2 x-8$, that is, $x=6$. This value fits: $a=\log _{\sqrt{4}} 2=1, c=\log _{2^{2}} 4=1$.

If $b=1$, then $\sqrt{5 x-26}=2 x-8$. This equation is equivalent to the equation $4 x^{2}-37 x+90=0$, which has no roots, so the cases $a=b=1, c=2$ and $b=c=1, a=2$ do not fit.

Thus, $x=6$ is the only solution to the problem.