Chapter 9.6, Problem 39E

To determine

Expand the logarithmic expression.

Answer to Problem 39E

The expanded form of expression $\ln \frac{4x^6}{y}$ is $1.38+6\ln x-\ln y$ .

Explanation of Solution

Given information:

The expression $\ln \frac{4x^6}{y}$ .

Formula used:

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

  $\ln \frac{4x^6}{y}$

Applying the identity ${\log }_b\left(\frac{M}{N}\right)={\log }_{b}M-{\log }_{b}N$ ,

  $=\ln 4x^6-\ln y$

Applying the identity ${\log }_b\left(MN\right)={\log }_{b}M+{\log }_{b}N$ ,

  $=\ln 4+\ln x^6-\ln y$

Applying the identity ${\log }_b{\left(M\right)}^a=a{\log }_{b}M$ ,

  $=\ln 4+6\ln x-\ln y$

The value of $\ln 4$ is $1.38$

  $=1.38+6\ln x-\ln y$

The expanded form of expression $\ln \frac{4x^6}{y}$ is $1.38+6\ln x-\ln y$ .