Chapter 4.4, Problem 67E

To determine

To calculate: The expression to use the change of base formula , $In\left(x-\sqrt{x^2-1}\right)=In\left(x+\sqrt{x^2+1}\right)$

Answer to Problem 67E

The expended form is these both sides are equal.

Explanation of Solution

Given information:

Expend the expression, $In\left(x-\sqrt{x^2-1}\right)=In\left(x+\sqrt{x^2+1}\right)$

Formula used:

For change the base rule,

  ${\log }_{a}b=\frac{\log b}{\log a},aIn\left(x\right)=In{x}^a$

Calculation :

Consider the expression, $In\left(x-\sqrt{x^2-1}\right)=In\left(x+\sqrt{x^2+1}\right)$

Rewrite the expression,

1^st we solve the right part of the given equation.

  $-In\left(x-\sqrt{x^2-1}\right)=In{\left(x-\sqrt{x^2-1}\right)}^{-1}$

Apply the log base rule: ${\log }_{a}b=\frac{\log b}{\log a},aIn\left(x\right)=In{x}^a$

  $In{\left(x-\sqrt{x^2-1}\right)}^{-1}=In\left(\frac{1}{In\left(x-\sqrt{x^2-1}\right)}\right)$

Rationalize the denominator:

  $\begin{array}{c}In\left(\frac{1}{In\left(x-\sqrt{x^2-1}\right)}\cdot \frac{\left(x+\sqrt{x^2-1}\right)}{\left(x+\sqrt{x^2-1}\right)}\right)=In\left(\frac{\left(x+\sqrt{x^2-1}\right)}{x^2-x\sqrt{x^2-1}+x\sqrt{x^2-1}+{\left(\sqrt{x^2-1}\right)}^2}\right)\\ =\left[\frac{\left(x+\sqrt{x^2-1}\right)}{x^2-x^2+1}\right]\\ =In\left(x+\sqrt{x^2-1}\right)\end{array}$

The above expression can’t be simplified further.

The result is as: $$

  $\left(x+\sqrt{x^2-1}\right)$