2A. The positive real numbers satisfy the equation

$$
\log \left(1+a^{2}\right)-\log a-2 \log 2=1-\log \left(100+b^{2}\right)+\log b
$$

Calculate the sum $a+b$.
Solution. By algebraic transformations, from the properties of logarithms we get

$$
\begin{aligned}
& \log \left(1+a^{2}\right)+\log \left(100+b^{2}\right)=\log a+\log 4+\log 10+\log b \\
& \log \left[\left(1+a^{2}\right)\left(100+b^{2}\right)\right]=\log 40 a b
\end{aligned}
$$

Since $y=\log x$ is a monotonic function, we have

$$
\begin{gathered}
\log \left[\left(1+a^{2}\right)\left(100+b^{2}\right)\right]=\log 40 a b \\
\left(1+a^{2}\right)\left(100+b^{2}\right)=40 a b
\end{gathered}
$$

Given that $a, b \in \mathbb{R}^{+}$, according to the inequality between the arithmetic and geometric means, we have

$$
\begin{aligned}
& 1+a^{2} \geq 2 a \\
& 100+b^{2} \geq 20 b
\end{aligned}
$$

and the equality holds if and only if $a=1$ and $b=10$. Therefore, if $a \neq 1$ or $b \neq 10$, then

$$
\left(1+a^{2}\right)\left(100+b^{2}\right)>40 a b
$$

which contradicts $(*)$.

Thus, $a=1, b=10$, from which we get $\quad a+b=1+10=11$.