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 $k$.
 A: First we convert all of the logarithms to base $10$ (the base doesnt matter, they just should all be in the same base.) Then we cancel out the $\log a$ on the top and bottom and simplify the expression.
$$\large\frac{\log_ba}{\log_ca}=\frac{\frac{\log a}{\log b}}{\frac{\log a}{\log c}}=\frac{\frac{1}{\log b}}{\frac{1}{\log c}}=\frac{\log c}{\log b}=\log_bc=\frac{19}{99}$$
This means that $c=b^{19/99}$ by definition. That means that we can replace $c$ with $b^{19/99}$ in the formula $\frac{b}{c}$. 
$$\Large\frac{b}{c}=\frac{b}{b^{19/99}}=\frac{b^{99/99}}{b^{19/99}}=b^{\frac{99-19}{99}}=b^{\frac{80}{99}}$$
Therefore we can take $c^k=\frac{b}{c}$ and replace all of the parts and get
$$\Large (b^\frac{19}{99})^k=(b^\frac{19k}{99})=b^{\frac{80}{99}}$$
Since the base b is the same, then $19k$ must equal $80$, and $k=80/19$
A: Since $\log_p q =\dfrac{1}{\log_q p}$, we get
$$
\frac{19}{99}=\frac{\log_ba}{\log_ca}=\frac{\log_a c}{\log_a b} = \log_b c,
$$
so
$$
b^{19/99} = c.
$$
Hence
$$
c^{99/19} = b,
$$
so
$$
c^{80/19} = \frac b c.
$$
A: 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}
