# How do you use the properties of logarithms to expand ln((x^2-1)/x^3)?

Nov 19, 2017

$\ln \left(\frac{{x}^{2} - 1}{x} ^ 3\right) = \ln \left(x - 1\right) + \ln \left(x + 1\right) - 3 \ln \left(x\right)$

#### Explanation:

If $a , b > 0$ then:

$\ln \left(\frac{a}{b}\right) = \ln \left(a\right) - \ln \left(b\right)$

and:

$\ln \left(a b\right) = \ln \left(a\right) + \ln \left(b\right)$

Hence we also find:

ln(a^n) = ln(overbrace(a * a * ... * a)^"n times") = overbrace(ln(a) + ln(a) + ... + ln(a))^"n terms" = n ln(a)

So:

$\ln \left(\frac{{x}^{2} - 1}{x} ^ 3\right) = \ln \left({x}^{2} - 1\right) - \ln \left({x}^{3}\right)$

$\textcolor{w h i t e}{\ln \left(\frac{{x}^{2} - 1}{x} ^ 3\right)} = \ln \left(\left(x - 1\right) \left(x + 1\right)\right) - 3 \ln \left(x\right)$

$\textcolor{w h i t e}{\ln \left(\frac{{x}^{2} - 1}{x} ^ 3\right)} = \ln \left(x - 1\right) + \ln \left(x + 1\right) - 3 \ln \left(x\right)$