Logarithm Fraction Contest Math Question

Question

The question is as follows:

If $\dfrac{\log_ba}{\log_ca}=\dfrac{19}{99}$ then $\dfrac{b}{c}=c^k$. Compute $k$.

Answer 1

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 $k=80/19$

Answer 2

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 $$ c^{80/19} = \frac b c. $$

Answer 3

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}