Single Variable Calculus, Chapter 7, 7.2-2, Section 7.2-2, Problem 4

Expand the quantity $\ln \frac{3x^2}{(x + 1)^5}$ by using the Laws of Logarithm.


$
\begin{equation}
\begin{aligned}

\ln \frac{3x^2}{(x + 1)^5} =& \ln 3x^2 - \ln (x + 1)^5
&& \text{(recall that } \ln \frac{x}{y} = \ln x - \ln y)
\\
\\
\ln \frac{3x^2}{(x + 1)^5} =& \ln 3x^2 - 5 \ln (x + 1)
&& \text{(recall that } \ln (x)^k = k \ln x)
\\
\\
\ln \frac{3x^2}{(x + 1)^5} =& \ln (3) + \ln (x^2) - 5 \ln (x + 1)
&& \text{(recall that } \ln (xY) = \ln x + \ln y)
\\
\\
\ln \frac{3x^2}{(x + 1)^5} =& \ln (3) + 2 \ln x - 5 \ln (x + 1)
&& \text{(recall that } \ln (x)^k = k \ln x


\end{aligned}
\end{equation}
$