# How do you condense ln(x+1)+ln(x-1)-3ln(x) ?

Apr 17, 2016

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

#### Explanation:

Condense the last $\log$ by recognising that a coefficient outside of the log is the same as an exponent inside, or

$3 \ln x = \ln {x}^{3}$.

Using the law that $\log a + \log b = \log a b$, and $\log a - \log b = \log \left(\frac{a}{b}\right)$, then,

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

and

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

which you could also condense, using laws of exponents, into

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