Question

Write each expression as a sum and/or difference of logarithms. Express powers as factors.

$\ln {\left[\frac{x^2-x-2}{{\left(x+4\right)}^2}\right]}^{\frac{1}{3}}$

Short answer

The result is$\frac{1}{3}\left[\ln \left(x-2\right)+\ln \left(x+1\right)-2\ln \left(x+4\right)\right]$.

Step-by-step solution

Step 1. Given information.

Consider the given information.

$\ln {\left[\frac{x^2-x-2}{{\left(x+4\right)}^2}\right]}^{\frac{1}{3}}$

Step 2. Formula used.

The logarithm properties are defined as,

$\begin{array}{l}{\log }_a\left(\frac{M}{N}\right)={\log }_{a}M-{\log }_{a}N\\ {\log }_a\left(MN\right)={\log }_{a}M+{\log }_{a}N\\ {\log }_a{M}^r=r{\log }_{a}M\end{array}$

Step 3. Calculation.

Apply the logarithm property to simplify.

$\begin{array}{rcl}\ln {\left[\frac{x^2-x-2}{{\left(x+4\right)}^2}\right]}^{\frac{1}{3}}& =& \frac{1}{3}\ln \left[\frac{x^2-x-2}{{\left(x+4\right)}^2}\right]\\ & =& \frac{1}{3}\left[\ln \left(x^2-x-2\right)-\ln {\left(x+4\right)}^2\right]\\ & =& \frac{1}{3}\left[\ln \left(x-2\right)\left(x+1\right)-2\ln \left(x+4\right)\right]\\ & =& \frac{1}{3}\left[\ln \left(x-2\right)+\ln \left(x+1\right)-2\ln \left(x+4\right)\right]\end{array}$