5. [5 points] Given the numbers $\log _{\left(\frac{x}{2}-1\right)^{2}}\left(\frac{x}{2}-\frac{1}{4}\right), \log _{\sqrt{x-\frac{11}{4}}}\left(\frac{x}{2}-1\right), \log _{\frac{x}{2}-\frac{1}{4}}\left(x-\frac{11}{4}\right)^{2}$. For which $x$ are two of these numbers equal, and the third one greater than them by 1?
Answer: $x=5$.

Solution. From the condition, it follows that the functions $\frac{x}{2}-1, \frac{x}{2}-\frac{1}{4}, x-\frac{11}{4}$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\sqrt{x-\frac{11}{4}}}\left(\frac{x}{2}-1\right), b=\log _{\frac{x}{2}-\frac{1}{4}}\left(x-\frac{11}{4}\right)^{2}, c=$ $\log _{\left(\frac{x}{2}-1\right)^{2}}\left(\frac{x}{2}-\frac{1}{4}\right)$. Then

$$
\begin{aligned}
a b c=\log _{\sqrt{x-\frac{11}{4}}}\left(\frac{x}{2}-1\right) \cdot \log _{\frac{x}{2}-\frac{1}{4}}\left(x-\frac{11}{4}\right)^{2} & \cdot \log _{\left(\frac{x}{2}-1\right)^{2}}\left(\frac{x}{2}-\frac{1}{4}\right)= \\
& =2 \log _{x-\frac{11}{4}}\left(\frac{x}{2}-1\right) \cdot 2 \log _{\frac{x}{2}-\frac{1}{4}}\left(x-\frac{11}{4}\right) \cdot \frac{\log _{x-\frac{11}{4}}\left(\frac{x}{2}-\frac{1}{4}\right)}{2 \log _{x-\frac{11}{4}}\left(\frac{x}{2}-1\right)}=2
\end{aligned}
$$

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

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

Consider case (13). Substituting $b=a$ and $c=a+1$ into the equation $a b c=2$ obtained above, 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 (14) and (15) 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 $\frac{x}{2}-\frac{1}{4}=x-\frac{11}{4}$, that is, $x=5$. This value fits: $c=\log _{\frac{9}{4}} \frac{9}{4}=1, a=\log _{\frac{3}{2}} \frac{3}{2}=1$.

If $b=1$, then $\frac{x}{2}-\frac{1}{4}=\left(x-\frac{11}{4}\right)^{2}$. This equation is equivalent to the equation $x^{2}-6 x+\frac{125}{16}=0$, which has roots $x=3 \pm \frac{\sqrt{19}}{4}$. The value $x=3-\frac{\sqrt{19}}{4}$ does not satisfy the domain of admissible values (since $x<\frac{11}{4}$). If $x=3+\frac{\sqrt{19}}{4}$, then $a=\log _{\sqrt{\frac{1+\sqrt{19}}{4}}}\left(\frac{4+\sqrt{19}}{8}\right)$, which is different from 1 and 2. Therefore, the cases $a=b=1, c=2$ and $b=c=1, a=2$ do not fit.

Thus, $x=5$ is the only solution to the problem.