<s>[INST] Logarithm Fraction Contest Math Question The question is as follows:
If $\dfrac{\log_ba}{\log_ca}=\dfrac{19}{99}$ then $\dfrac{b}{c}=c^k$. Compute $k$.
 [/INST] Rewrite
$$
\begin{align}
\dfrac{\log_ba}{\log_ca}&=\dfrac{19}{99}\\
\dfrac{\frac{\log a}{\log b}}{\frac{\log a}{\log c}}&=\dfrac{19}{99}\\
\dfrac{\log a}{\log b}\cdot\dfrac{\log c}{\log a}&=\dfrac{19}{99}\\
\dfrac{\log c}{\log b}&=\dfrac{19}{99}\\
99\log c&=19\log b\\
\log c^{99}&=\log b^{19}\\
c^{99}&=b^{19}\tag1
\end{align}
$$
and
$$
\frac{b}{c}=c^k\quad\Rightarrow\quad b=c^{k+1}\quad\Rightarrow\quad b^{19}=c^{19(k+1)}.\tag2
$$
Substituting $(1)$ to $(2)$ yields
\begin{align}
c^{99}=c^{19(k+1)}\quad\Rightarrow\quad 19(k+1)&=99\\
k+1&=\frac{99}{19}\\
k&=\frac{99}{19}-1\\
&=\large\color{blue}{\frac{80}{19}}.
\end{align}
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