2. Determine all real numbers $a$ and $b (a < b)$ for which

$$
|\lg (a+1)|=\left|\lg \left(-\frac{b+1}{b+2}+1\right)\right| \text { and }|\lg (10 a+6 b+22)|=4 \lg 2 .
$$
Solution. Clearly, $0<a+1$, and according to the condition of the problem, the inequalities $0<a+1<b+1<b+2$ are correct. Furthermore, from $|\lg (a+1)|=\left|\lg \left(-\frac{b+1}{b+2}+1\right)\right|$ it follows that

$$
|\lg (a+1)|=\left|\lg \left(-\frac{b+1}{b+2}+1\right)\right|=\left|\lg \frac{1}{b+2}\right|=|-\lg (b+2)|=|\lg (b+2)|,
$$

so we have two cases.

Case 1. If $a+1=b+2$, i.e., $a=b+1$, then from the second condition it follows that

$$
|\lg (16 b+32)|=\lg 16
$$

so $16 b+32=16$ or $16 b+32=\frac{1}{16}$, i.e., $b_{1}=-1$ or $b_{2}=-2+\frac{1}{256}$, but both solutions contradict the condition $0<b+1$.

Case 2. If $a+1=\frac{1}{b+2}$, then from the second condition it follows that

$$
\left|\lg \left(6(b+2)+\frac{10}{b+2}\right)\right|=\lg 16
$$

so $6(b+2)+\frac{10}{b+2}=16$ or $6(b+2)+\frac{10}{b+2}=\frac{1}{16}$. We introduce the substitution $b+2=u$. In the first case, we get $6 u^{2}-16 u+10=0$, so $u_{1}=1$ and $u_{2}=\frac{10}{6}$, i.e., $b_{1}=-1$ and $b_{2}=-\frac{1}{3}$. The solution $b=-1$ does not satisfy the condition $0<b+1$, and for $b=-\frac{1}{3}$ we get $a=-\frac{2}{5}$. In the second case, we get the equation $96 u^{2}-u+160=0$ which has no real solutions.

Finally, $a=-\frac{2}{5}$ and $b=-\frac{1}{3}$ are the only real numbers that satisfy the given equalities.