Chapter 4 - Section 4.4 - Laws of Logarithms - 4.4 exercises - Page 359: 71

See proof below

Prove that $-\ln (x - \sqrt{x^2 - 1} = \ln (x+ \sqrt {x^2 - 1})$ $-\ln (x - \sqrt{x^2 - 1} = \ln ((x - \sqrt{x^2 - 1}) ^ {-1})$ $= \ln (\frac{1}{x - \sqrt{x^2 - 1}})$ $* \frac{x+ \sqrt {x^2 - 1}}{x+ \sqrt {x^2 - 1}}$ $= \ln (\frac{x+ \sqrt {x^2 - 1}}{x^2 - x^2 + 1})$ $= \ln (\frac{x+ \sqrt {x^2 - 1}}{1})$ $ = \ln(x+ \sqrt {x^2 - 1})$