# How to express this equation in terms of m and n ?

## Given ${\log}_{8} 3 = m$ and ${\log}_{8} 5 = n$, express ${\log}_{3} 50$ in terms of m and n.

Sep 4, 2017

${\log}_{3} \left(50\right) = \frac{1 + 6 n}{3 m}$

#### Explanation:

We need to know the logarithm rules:

• ${\log}_{a} \left(b c\right) = {\log}_{a} \left(b\right) + {\log}_{a} \left(c\right)$
• ${\log}_{a} \left({b}^{c}\right) = c {\log}_{a} \left(b\right)$
• ${\log}_{a} \left(b\right) = {\log}_{c} \frac{b}{\log} _ c \left(a\right)$
• ${\log}_{a} \left(a\right) = 1$

Then:

${\log}_{3} \left(50\right) = {\log}_{8} \frac{50}{\log} _ 8 \left(3\right) = {\log}_{8} \frac{50}{m}$

Splitting up $50$ into its prime factorization:

$= {\log}_{8} \frac{2 \cdot {5}^{2}}{m} = \frac{{\log}_{8} \left(2\right) + {\log}_{8} \left({5}^{2}\right)}{m} = \frac{{\log}_{8} \left({8}^{\frac{1}{3}}\right) + {\log}_{8} \left({5}^{2}\right)}{m}$

$= \frac{\frac{1}{3} {\log}_{8} \left(8\right) + 2 {\log}_{8} \left(5\right)}{m} = \frac{\frac{1}{3} + 2 n}{m} = \frac{1 + 6 n}{3 m}$