Chapter 6.5, Problem 18E

To determine

To calculate: The expanded form of the expression $\ln \frac{6x^2}{y^4}$ .

Answer to Problem 18E

The required expanded form of the expression $\ln \frac{6x^2}{y^4}$ is $\ln 6+2\ln x-4\ln y$ .

Explanation of Solution

Given information:

The given expression is $\ln \frac{6x^2}{y^4}$ .

Formula used:

Quotient property of logarithm:

  ${\log }_b\frac{m}{n}={\log }_{b}m-{\log }_{b}n$

Product property of logarithm:

  ${\log }_{b}mn={\log }_{b}m+{\log }_{b}n$

Power property of logarithm:

  ${\log }_b{m}^n=n{\log }_{b}m$

Calculation:

Consider the expression $\ln \frac{6x^2}{y^4}$ .

Expand the expression using quotient property of logarithm ${\log }_b\frac{m}{n}={\log }_{b}m-{\log }_{b}n$ .

  $\ln \frac{6x^2}{y^4}=\ln 6x^2-\ln y^4$

Expand the above expression using product property of logarithm ${\log }_{b}mn={\log }_{b}m+{\log }_{b}n$ .

  $\begin{array}{c}\ln \frac{6x^2}{y^4}=\ln 6x^2-\ln y^4\\ =(\ln 6+\ln x^2)-\ln y^4\end{array}$

Expand the above expression using power property of logarithm ${\log }_b{m}^n=n{\log }_{b}m$ .

  $\begin{array}{c}\ln \frac{6x^2}{y^4}=(\ln 6+\ln x^2)-\ln y^4\\ =\ln 6+2\ln x-4\ln y\end{array}$

Hence, the required expanded form of the expression $\ln \frac{6x^2}{y^4}$ is $\ln 6+2\ln x-4\ln y$ .