# How do you approximate log_3 (2/3) given log_3 2=0.6310 and log_3 7=1.7712?

Nov 1, 2016

${\log}_{3} \left(\frac{2}{3}\right) = - 0.369$

#### Explanation:

Using the difference rule of logarithms, ${\log}_{a} \left(\frac{n}{m}\right) = {\log}_{a} n - {\log}_{a} m$, we can rewrite ${\log}_{3} \left(\frac{2}{3}\right)$ as ${\log}_{3} \left(2\right) - {\log}_{3} \left(3\right)$

We know that ${\log}_{3} 2 = 0.6310$.

$\implies 0.6310 - {\log}_{3} 3$

Let's use the change of base rule to see what ${\log}_{3} 3$ can be simplified to.

${\log}_{3} \left(3\right) = \log \frac{3}{\log} 3 = 1$

So, ${\log}_{3} \left(\frac{2}{3}\right) = 0.6310 - 1 = - 0.369$

Hopefully this helps!