# How do you approximate log_5 30 given log_5 2=0.4307 and log_5 3=0.6826?

Nov 29, 2016

Rewrite $30$ using multiples of $2$, $3$, and $5$.

${\log}_{5} \left(30\right) = {\log}_{5} \left(2 \cdot 3 \cdot 5\right)$

Using the rule ${\log}_{a} \left(b c\right) = {\log}_{a} \left(b\right) + {\log}_{a} \left(c\right)$, this becomes

${\log}_{5} \left(30\right) = {\log}_{5} \left(2\right) + {\log}_{5} \left(3\right) + {\log}_{5} \left(5\right)$

Using ${\log}_{2} \left(5\right) \approx 0.4307$, ${\log}_{3} \left(5\right) \approx 0.6828$, and the rule that ${\log}_{a} \left(a\right) = 1$:

${\log}_{5} \left(30\right) \approx 0.4307 + 0.6826 + 1 \approx 2.1133$