Chapter 4.4, Problem 36E

To determine

To calculate: The expression to use the laws of logarithm , ${\log }_5\sqrt{\frac{x-1}{x+1}}$

Answer to Problem 36E

The expended form is, $0$

Explanation of Solution

Given information:

Expend the expression, ${\log }_5\sqrt{\frac{x-1}{x+1}}$

Formula used:

For change the base rule,

  $\begin{array}{l}{\log }_{a}b=\frac{\log b}{\log a},{\log }_a\left(\frac{x^t}{y{z}^{2t}}\right)={\log }_a{x}^t-{\log }_{a}y-{\log }_a{z}^3\\ {\log }_a{b}^x=x{\log }_{a}b\end{array}$

Calculation :

Consider the expression, ${\log }_5\sqrt{\frac{x-1}{x+1}}$

Rewrite the expression,

  $\begin{array}{c}{\log }_5\sqrt{\frac{x-1}{x+1}}=\frac{{\log }_5\sqrt{x-1}}{{\log }_5\sqrt{x+1}}\\ ={\log }_5\sqrt{x-1}-{\log }_5\sqrt{x+1}\end{array}$

Apply the log base rule:

  $\begin{array}{l}\log ab=\log a+\log b,{\log }_a\left(\frac{x^t}{y{z}^{2t}}\right)={\log }_a{x}^t-{\log }_{a}y-{\log }_a{z}^3\\ {\log }_a{b}^x=x{\log }_{a}b\end{array}$

  $\begin{array}{c}{\log }_5\sqrt{x-1}-{\log }_5\sqrt{x+1}={\log }_5{\left(x-1\right)}^{\frac{1}{2}}-{\log }_5{\left(x+1\right)}^{\frac{1}{2}}\\ =\frac{1}{2}{\log }_5\left(x-1\right)+\frac{1}{2}{\log }_5\left(x+1\right)\\ =\frac{1}{2}{\log }_{5}x-\frac{1}{2}{\log }_{5}1+\frac{1}{2}{\log }_{5}x+\frac{1}{2}{\log }_{5}1\end{array}$

Here ${\log }_{e}1=0$

  $\begin{array}{c}\frac{1}{2}{\log }_{5}x-\frac{1}{2}{\log }_{5}1-\frac{1}{2}{\log }_{5}x+\frac{1}{2}{\log }_{5}1=\frac{1}{2}{\log }_{5}x-0-\frac{1}{2}{\log }_{5}x-0\\ =0\end{array}$

The above expression can’t be simplified further.

Thus the simplified form of the expression is $0$