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

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

$$
\begin{aligned}
& a b c=\log _{\sqrt{5 x-1}}(4 x+1) \cdot \log _{4 x+1}\left(\frac{x}{2}+2\right)^{2} \cdot \log _{\frac{x}{2}+2}(5 x-1)= \\
& \quad=2 \log _{5 x-1}(4 x+1) \cdot 2 \log _{4 x+1}\left(\frac{x}{2}+2\right) \cdot \frac{\log _{4 x+1}(5 x-1)}{\log _{4 x+1}\left(\frac{x}{2}+2\right)}=4
\end{aligned}
$$

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

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

Consider case (3). 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 (4) and (5) 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 $5 x-1=\frac{x}{2}+2$, that is, $x=\frac{2}{3}$. Therefore, $a=2 \log _{\frac{7}{3}} \frac{11}{3} \neq 2$, that is, there are no values of $x$ for which $a=b=2, c=1$.

If $a=1$, then $4 x+1=\sqrt{5 x-1}$. Squaring both sides of the last equation, we get the equation $16 x^{2}+3 x+2=0$, which has no roots, so the case $a=1, b=c=2$ also does not fit.

If $b=1$, then $\left(\frac{x}{2}+2\right)^{2}=4 x+1$. This equation is equivalent to the equation $x^{2}-8 x+12=0$, the roots of which are $x=2$ and $x=6$, but $x=6$ does not fit, because in this case $a=\log _{\sqrt{29}} 25 \neq 2$. The value $x=2$ fits: $a=\log _{\sqrt{9}} 9=2, c=\log _{3} 9=2$.

Thus, $x=2$ is the only solution to the problem.