## Task B-3.4.

The number of Klara's years is equal to $\log _{\sqrt[3]{5}} a^{2}$, the number of Maria's years is equal to $\log _{5 \sqrt{5}}\left(125 b^{9}\right)$, and the number of Jan's years is equal to $\frac{1}{\log _{c^{2}}^{\sqrt[3]{5}}}$. What is the sum of their ages if $a b c=625$?
## Solution.

Let's write the given logarithms in a simpler form:

$$
\begin{array}{cc}
\log _{\sqrt[3]{5}} a^{2}=2 \log _{5^{\frac{1}{3}}} a=6 \log _{5} a, & 1 \text { point } \\
\log _{5 \sqrt{5}}\left(125 b^{9}\right)=\log _{5^{\frac{3}{2}}} 125+\log _{5^{\frac{3}{2}}} b^{9}=3 \cdot \frac{2}{3} \log _{5} 5+9 \cdot \frac{2}{3} \log _{5} b=2+6 \log _{5} b, & 2 \text { points } \\
\frac{1}{\log _{c^{2}} \sqrt[3]{5}}=\frac{1}{\frac{1}{2} \log _{c} 5^{\frac{1}{3}}}=\frac{1}{\frac{1}{6} \log _{c} 5}=6 \log _{5} c . & 1 \text { point }
\end{array}
$$

The desired sum of years is equal to:

$6 \log _{5} a+2+6 \log _{5} b+6 \log _{5} c=2+6 \log _{5} a b c=2+6 \log _{5} 625=2+6 \cdot 4=26$ years.