## Task B-3.2.

Let $a$ and $b$ be positive real numbers such that $\frac{\log a}{3}=\frac{\log b}{2}$ and $\log (a b)=5$. What is $\sqrt[3]{a}+b^{\frac{1}{2}}$?
## First Solution.

First, we rewrite the initial equation in another form:

\[
\frac{\log a}{3} = \frac{\log b}{2} \cdot 6
\]

\[
2 \log a = 3 \log b
\]

\[
2 \log a - 3 \log b = 0
\]

By applying the properties of logarithms, we transform the second equation:

\[
\log (a b) = \log a + \log b = 5
\]

We obtain a system of two equations with two unknowns:
\[
\left\{
\begin{array}{l}
2 \log a - 3 \log b = 0 \\
\log a + \log b = 5
\end{array}
\right.
\]

Solving this system, we find that \(\log a = 3\) and \(\log b = 2\).

2 points

Thus, it follows that \(a = 10^3 = 1000\) and \(b = 10^2 = 100\). 1 point

Finally, \(\sqrt[3]{a} + b^{\frac{1}{2}} = \sqrt[3]{1000} + 100^{\frac{1}{2}} = 10 + 10 = 20\).