5. [5 points] Given the numbers $\log _{\left(\frac{x}{2}+1\right)^{2}}\left(\frac{7 x}{2}-\frac{17}{4}\right), \log _{\sqrt{\frac{7 x}{2}-\frac{17}{4}}}\left(\frac{3 x}{2}-6\right)^{2}, \log _{\sqrt{\frac{3 x}{2}-6}}\left(\frac{x}{2}+1\right)$. For which $x$ are two of these numbers equal, and the third one less than them by 1?
Answer: $x=7$.

Solution. From the condition, it follows that the functions $\left(\frac{x}{2}+1\right),\left(\frac{7 x}{2}-\frac{17}{4}\right),\left(\frac{3 x}{2}-6\right)$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\left(\frac{x}{2}+1\right)^{2}}\left(\frac{7 x}{2}-\frac{17}{4}\right), b=$ $\log _{\sqrt{\frac{7 x}{2}-\frac{17}{4}}}\left(\frac{3 x}{2}-6\right)^{2}, c=\log _{\sqrt{\frac{3 x}{2}-6}}\left(\frac{x}{2}+1\right)$. Then

$$
\begin{aligned}
a b c=\log _{\left(\frac{x}{2}+1\right)^{2}}\left(\frac{7 x}{2}-\frac{17}{4}\right) \cdot & \log _{\sqrt{\frac{7 x}{2}-\frac{17}{4}}}\left(\frac{3 x}{2}-6\right)^{2} \cdot \log _{\sqrt{\frac{3 x}{2}-6}}\left(\frac{x}{2}+1\right)= \\
& =\frac{1}{2} \log _{\left(\frac{x}{2}+1\right)}\left(\frac{7 x}{2}-\frac{17}{4}\right) \cdot 4 \log _{\frac{7 x}{2}-\frac{17}{4}}\left(\frac{3 x}{2}-6\right) \cdot 2 \frac{\log _{\frac{7 x}{2}-\frac{17}{4}}\left(\frac{x}{2}+1\right)}{\log _{\frac{7 x}{2}-\frac{17}{4}}\left(\frac{3 x}{2}-6\right)}=4
\end{aligned}
$$

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

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

Consider case (28). Substituting $b=a$ and $c=a-1$ into the equation $a b c=4$, 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 (29) and (30) 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 $c=2$, then $\frac{3 x}{2}-6=\frac{x}{2}+1$, that is, $x=7$. This value fits: $a=\log _{\left(\frac{9}{2}\right)^{2}} \frac{81}{4}=1, b=\log _{\sqrt{\frac{81}{4}}}\left(\frac{9}{2}\right)=2$. If $c=1$, then $\sqrt{\frac{3 x}{2}-6}=\frac{x}{2}+1$. From this equation, after squaring, we get that $x^{2}-2 x+28=0$, that is, there are no roots.

Thus, $x=7$ is the only solution to the problem.