# How do you evaluate log_9 (1/3)+ 3 log_9 3?

Mar 23, 2016

Let's first simplify with the rule $a \log x = \log {x}^{a}$.

#### Explanation:

${\log}_{9} \left(\frac{1}{3}\right) + {\log}_{9} \left({3}^{3}\right)$

=${\log}_{9} \left(\frac{1}{3}\right) + {\log}_{9} \left(27\right)$

Since we're in the same base, we can simplify using the rule ${\log}_{a} n + {\log}_{a} m = {\log}_{a} \left(n \times m\right)$

=${\log}_{9} \left(\frac{1}{3} \times 27\right)$

=${\log}_{9} \left(9\right)$

We can now evaluate using the rule ${\log}_{a} a = 1$

=$1$

Hopefully this helps!