College Algebra, Chapter 5, 5.3, Section 5.3, Problem 14

Evaluate the expression $\displaystyle \log_3 100 - \log_3 18 - \log_3 50$


$\begin{equation} \begin{aligned} \log_3 100 - \log_3 18 - \log_3 50 =& \log_3 100 - (\log_3 18 + \log_3 50) && \\ \\ \log_3 100 - \log_3 18 - \log_3 50 =& \log_3 100 - \log_3 (18 \cdot 50) && \text{Law of Logarithms } \log_a (AB) = \log_a A + \log_a B \\ \\ \log_3 100 - \log_3 18 - \log_3 50 =& \log_3 \left( \frac{100}{900} \right) && \text{Law of Logarithms } \log_a \left( \frac{A}{B} \right) = \log_a A - \log_a B \\ \\ \log_3 100 - \log_3 18 - \log_3 50 =& \log_3 \left( \frac{1}{9} \right) && \text{Simplify} \\ \\ \log_3 100 - \log_3 18 - \log_3 50 =& -2 && \text{Because } 3^{-2} = \frac{1}{9} \end{aligned} \end{equation}$