Chapter 5.5, Problem 53AYU

To determine

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

Answer to Problem 53AYU

  $\boxed{\ln {\left(\frac{x^2-x-2}{{\left(x+4\right)}^2}\right)}^{1/3}=\frac{1}{3}\ln \left(x+1\right)+\frac{1}{3}\ln \left(x-2\right)-\frac{2}{3}\ln \left(x+4\right)}$

Explanation of Solution

Given information:

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

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

Calculation:

Consider the given expression,

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

Apply logarithm property.

  ${\log }_a{M}^r=r{\log }_{a}M$

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

Now factorise $x^2-x-2$

  $\begin{array}{l}{x}^2-x-2=\left(x+1\right)\left(x-2\right)\\ now,\\ \ln {\left(\frac{x^2-x-2}{{\left(x+4\right)}^2}\right)}^{1/3}=\frac{1}{3}\ln \left(\frac{\left(x+1\right)\left(x-2\right)}{{\left(x+4\right)}^2}\right)\\ now,Apply\\ {\log }_a\left(\frac{M}{N}\right)={\log }_{a}M-{\log }_{a}N\\ \frac{1}{3}\ln \left(\frac{\left(x+1\right)\left(x-2\right)}{{\left(x+4\right)}^2}\right)=\frac{1}{3}\ln \left[\left(x+1\right)\left(x-2\right)\right]-\frac{1}{3}\ln {\left(x+4\right)}^2\\ \frac{1}{3}\ln \left(\frac{\left(x+1\right)\left(x-2\right)}{{\left(x+4\right)}^2}\right)=\frac{1}{3}\ln \left(x+1\right)+\frac{1}{3}\ln \left(x-2\right)-\frac{2}{3}\ln \left(x+4\right)\end{array}$

  $\boxed{\ln {\left(\frac{x^2-x-2}{{\left(x+4\right)}^2}\right)}^{1/3}=\frac{1}{3}\ln \left(x+1\right)+\frac{1}{3}\ln \left(x-2\right)-\frac{2}{3}\ln \left(x+4\right)}$

Hence, $\boxed{\ln {\left(\frac{x^2-x-2}{{\left(x+4\right)}^2}\right)}^{1/3}=\frac{1}{3}\ln \left(x+1\right)+\frac{1}{3}\ln \left(x-2\right)-\frac{2}{3}\ln \left(x+4\right)}$