## Task B-4.7.

Solve the inequality

$$
\begin{aligned}
& \log _{\sqrt{5}}\left(3^{x^{2}-x-1}+2\right)+\log _{5}\left(3^{x^{2}-x-1}+2\right)+\log _{25}\left(3^{x^{2}-x-1}+2\right)+ \\
& \log _{625}\left(3^{x^{2}-x-1}+2\right)+\ldots+\log _{5^{2}}\left(3^{x^{2}-x-1}+2\right)+\ldots \leqslant 4
\end{aligned}
$$
## Solution.

Let $3^{x^{2}-x-1}+2=y$.

The given inequality is successively equivalent to

$$
\begin{gathered}
\log _{\sqrt{5}} y+\log _{5} y+\log _{25} y+\log _{625} y+\ldots+\log _{5^{2 k}} y+\ldots \leqslant 4 \\
2 \log _{5} y+\log _{5} y+\frac{1}{2} \log _{5} y+\frac{1}{4} \log _{5} y+\ldots+\frac{1}{2^{k}} \log _{5} y+\ldots \leqslant 4 \\
\left(2+1+\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{k}}+\ldots\right) \log _{5} y \leqslant 4
\end{gathered}
$$

The expression in parentheses is a geometric series with the first term 2 and the common ratio $\frac{1}{2}$,

so its sum is $2+1+\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{k}}+\ldots=\frac{2}{1-\frac{1}{2}}=4$.

It follows that $4 \log _{5} y \leqslant 4$, or $\log _{5}\left(3^{x^{2}-x-1}+2\right) \leqslant 1$

and further we have

$$
\begin{array}{cc}
0<3^{x^{2}-x-1}+2 \leqslant 5 & 1 \text { point } \\
3^{x^{2}-x-1} \leqslant 3 & \\
x^{2}-x-1 \leqslant 1 & 1 \text { point } \\
(x-2)(x+1) \leqslant 0 &
\end{array}
$$

The final solution is $x \in[-1,2]$.