# How do you evaluate log_3 (1/81)?

Oct 18, 2016

By reducing the logarithm using the laws of logs, and reducing the number to a power of the base the result follows. $= - 4$

#### Explanation:

$\textcolor{red}{{\log}_{\text{3"(1/81)=log_"3"(1)-log_"3}} 81}$

since

${\log}_{a} \left(\frac{X}{Y}\right) = {\log}_{a} \left(X\right) - {\log}_{a} \left(Y\right)$

$\textcolor{red}{{\log}_{\text{3"(1/81)=0-log_"3}} {3}^{4}}$

since ${\log}_{a} 1 = 0$

$\forall a \in \mathbb{R}$

$\textcolor{red}{{\log}_{\text{3"(1/81)=-4log_"3}} 3}$

since ${\log}_{a} {X}^{n} = n {\log}_{a} X$

$\textcolor{red}{{\log}_{\text{3}} \left(\frac{1}{81}\right) = - 4}$

since ${\log}_{a} a = 1$