Task B-3.2. Solve the inequality:

$$
\log _{\frac{x+4}{2}} \log _{2} \frac{2 x-1}{1+x}<0
$$
## Solution.

Determine the conditions for which the expression on the left side of the inequality is defined:

$$
\begin{aligned}
& \frac{2 x-1}{1+x}>0 \\
& \log _{2} \frac{2 x-1}{1+x}>0, \text { i.e., } \frac{2 x-1}{1+x}>1 \\
& \frac{x+4}{2}>0, \frac{x+4}{2} \neq 1, \text { i.e., } x>-4, x \neq-2
\end{aligned}
$$

From (1) and (2), it follows that $\frac{2 x-1}{1+x}>1$. This implies $\frac{x-2}{1+x}>0$, or $x \in\langle-\infty,-1\rangle \cup\langle 2, \infty\rangle$, which together with (3) gives the final condition $x \in(\langle-4,-1\rangle \cup\langle 2, \infty\rangle) \backslash\{-2\}$.

Next, solving the inequality is divided into two cases, when the base is $0<\frac{x+4}{2}<1$ and when the base is $\frac{x+4}{2}>1$.

In the first case $0<\frac{x+4}{2}<1$ (i.e., for $x \in (-4, -2)$), we have:

$$
\begin{aligned}
\log _{2} \frac{2 x-1}{1+x} & >\log _{2}\left(\frac{x+4}{2}\right)^{0} \\
\frac{2 x-1}{1+x} & >2 \\
\frac{-3}{1+x} & >0
\end{aligned}
$$

Then $x+1<0$, i.e., $x<-1$.

In the second case $\frac{x+4}{2}>1$ (i.e., for $x>-2$), we have:

$$
\begin{aligned}
\log _{2} \frac{2 x-1}{1+x} & >1 \\
\frac{2 x-1}{1+x} & >2 \\
\frac{-3}{1+x} & >0
\end{aligned}
$$

Then $x+1<0$, i.e., $x<-1$.

Finally, the solution is the union of the solutions of the two previous cases, intersected with the initial conditions. This is $x \in\langle-4,-2\rangle \cup\langle 2, \infty\rangle$.