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

Solution. From the condition, it follows that the functions $x+1, 2x-3$ are positive and do not take the value 1 for all $x$ in the domain of admissible values. Let $a=\log _{\sqrt{2 x-3}}(x+1), b=\log _{2 x^{2}-3 x+5}(2 x-3)^{2}, c=\log _{x+1}\left(2 x^{2}-3 x+5\right)$. Then

$$
\begin{aligned}
a b c=\log _{\sqrt{2 x-3}}(x+1) \cdot & \log _{2 x^{2}-3 x+5}(2 x-3)^{2} \cdot \log _{x+1}\left(2 x^{2}-3 x+5\right)= \\
& =2 \log _{2 x-3}(x+1) \cdot 2 \log _{2 x^{2}-3 x+5}(2 x-3) \cdot \frac{\log _{2 x-3}\left(2 x^{2}-3 x+5\right)}{\log _{2 x-3}(x+1)}=4
\end{aligned}
$$

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

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

Consider case (23). 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 (24) and (25) 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 $x+1=2 x^{2}-3 x+5$. This equation has no roots, so there are no values of $x$ for which $a=b=2, c=1$.

If $b=1$, then $2 x^{2}-3 x+5=(2 x-3)^{2}$. This equation is equivalent to the equation $2 x^{2}-9 x+4=0$, the roots of which are $x=\frac{1}{2}$, which does not satisfy the domain of admissible values, and $x=4$. The value $x=4$ is valid: $a=\log _{\sqrt{5}} 5=2$, $c=\log _{5} 25=2$.

If $a=1$, then $\sqrt{2 x-3}=x+1$. Squaring both sides of the equation, we get the equation $x^{2}=-4$, which has no roots. Therefore, there are no values of $x$ for which $b=c=2, a=1$.

Thus, $x=4$ is the only solution to the problem.