Monday, April 22, 2013

College Algebra, Chapter 5, 5.3, Section 5.3, Problem 50

Combine the expression $\displaystyle \ln (a + b) + \ln (a - b) - 2 \ln c$, using the Laws of Logarithm


$
\begin{equation}
\begin{aligned}

\ln (a + b) + \ln (a - b) - 2 \ln c =& \ln (a + b) + \ln (a - b) - \ln c^2
&& \text{Law of Logarithm } \log_a (A^C) = C \log_a A
\\
\\
\ln (a + b) + \ln (a - b) - 2 \ln c =& \ln (a + b)(a - b) - \ln c^2
&& \text{Law of Logarithm } \log_a (AB) = \log_a A + \log_a B
\\
\\
\ln (a + b) + \ln (a - b) - 2 \ln c =& \ln \left[ \frac{(a + b)(a - b)}{c^2} \right]
&& \text{Law of Logarithm } \log_a \left( \frac{A}{B} \right) = \log_a A - \log_a B
\\
\\
\ln (a + b) + \ln (a - b) - 2 \ln c =& \ln \left( \frac{a^2 - b^2}{c^2} \right)
&& \text{Special Product } (A - B)(A + B) = (A^2 - B^2)

\end{aligned}
\end{equation}
$

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