## Task B-3.2.

If $\log _{2} 3=a$ and $\log _{7} 2=b$, what is $\log _{6} 28$?
## Solution.

$$
\begin{array}{rlr}
\log _{6} 28 & =\frac{\log _{2} 28}{\log _{2} 6} & 1 \text { point } \\
& =\frac{\log _{2}(4 \cdot 7)}{\log _{2}(2 \cdot 3)}=\frac{\log _{2} 4+\log _{2} 7}{\log _{2} 2+\log _{2} 3} & 2 \text { points } \\
& =\frac{2+\frac{1}{\log _{7} 2}}{1+\log _{2} 3} & 2 \text { points } \\
& =\frac{2+\frac{1}{b}}{1+a}=\frac{2+b+1}{b+a b} & 1 \text { point }
\end{array}
$$