## Task B-3.1.

Simplify the numerical expression $\frac{6-\log _{14} 2401}{4-\log _{14} 49}$ and write it in the form of a logarithm $\log _{28} n$, where $n$ is a natural number.
## First solution.

First, note that $49=7^{2}$ and $2401=7^{4}$.

Applying the properties of logarithms, we simplify the given numerical expression.

$$
\begin{aligned}
\frac{6-\log _{14} 2401}{4-\log _{14} 49} & =\frac{6-4 \log _{14} 7}{4-2 \log _{14} 7}=\frac{3-2 \log _{14} 7}{2-\log _{14} 7} & & 1 \text { point } \\
& =\frac{\log _{14} 14^{3}-\log _{14} 7^{2}}{\log _{14} 14^{2}-\log _{14} 7} & & 1 \text { point } \\
& =\frac{\log _{14} \frac{14^{3}}{7^{2}}}{\log _{14} \frac{14^{2}}{7}} & & 1 \text { point } \\
& =\frac{\log _{14} 56}{\log _{14} 28} & & 1 \text { point } \\
& =\log _{28} 56 & & 1 \text { point }
\end{aligned}
$$