2. Find the number of roots of the equation: $2^{\lg \left(x^{2}-2023\right)}-\lg 2^{x^{2}-2022}=0$.
# Answer: 4 roots.

Solution: Using the properties of logarithms, rewrite the equations in the following form

$$
\left(x^{2}-2023\right)^{\lg 2}-\lg 2^{x^{2}-2022}=0
$$

Introduce the notations $z=x^{2}-2023, a=1 \mathrm{~g} 2$, in this case $z>0, a \in(0,1)$. Then $z^{a}=(z+1) a$

Let $y_{1}(z)=z^{a}, y_{2}(z)=(z+1) a$.

Since $y_{1}(1)=1, y_{2}(1)=2 a$, and $y_{1}(1)=1=\lg 10>\lg 4=2 \lg 2=2 a$ and considering the monotonicity and convexity of the functions $y_{1}(z), y_{2}(z)$ for $a \in(0,1)$, we get that the equation $z^{a}=(z+1) a$ has two roots $z_{1}$ and $z_{2}$, one of which, for example $z_{1}$ is less than one but greater than zero, and the other root $z_{2}$ will be greater than one. Then, recalling the substitution $z=x^{2}-2023$ and returning to the original variable $x$, we conclude that the original equation will have 4 roots: $\pm \sqrt{z_{1}+2023}, \pm \sqrt{z_{2}+2023}$.