Chapter 3.3, Problem 68E

To determine

To condense: The given logarithm.

Answer to Problem 68E

  $\frac{1}{5}\left[\ln x-\ln \left(x+1\right)-\ln \left(x-1\right)\right]$ can be condensed as $\ln \sqrt[5]{\frac{x}{x^2-1}}$ .

Explanation of Solution

Given:

The following logarithm:

  $\frac{1}{5}\left[\ln x-\ln \left(x+1\right)-\ln \left(x-1\right)\right]$

Formula used:

The following power property:

  $\ln u^n=n\ln u$

The following quotient property:

  $\ln \left(\frac{u}{v}\right)=\ln u-\ln v$

The following expansion formula:

  $a^2-b^2=\left(a+b\right)\left(a-b\right)$

Calculation:

The given logarithm can be condensed as follows:

  $\begin{array}{l}\frac{1}{5}\left[\ln x-\ln \left(x+1\right)-\ln \left(x-1\right)\right]=\frac{1}{5}\left[\ln \frac{x}{x+1}-\ln \left(x-1\right)\right][\text{Quotient property]}\\ =\frac{1}{5}\left[\ln \frac{\frac{x}{x+1}}{x-1}\right][\text{Quotient property]}\\ =\frac{1}{5}\left[\ln \frac{x}{\left(x+1\right)\left(x-1\right)}\right][\text{Simplify]}\\ =\frac{1}{5}\left[\ln \frac{x}{x^2-1}\right][\text{Expansion formula]}\\ =\ln {\left(\frac{x}{x^2-1}\right)}^{1/5}[\text{Power property]}\\ =\ln \sqrt[5]{\frac{x}{x^2-1}}[\text{Rewrite exponent as radical]}\end{array}$