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}
$