Simplify ln(x/(x-1))+ ln((x+1)/x)-ln(x^2-1) ?

Apr 9, 2017

$- 2 \log \left(x - 1\right)$

Explanation:

Using the logarithmic properties

$\log \left(a b\right) = \log a + \log b$
$\log {a}^{b} = b \log a$

we have

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

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

Apr 9, 2017

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

Addition of logarithms is the same as multiplication of their arguments:

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

Simplify a bit:

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

Subtractions of logarithms is the same as division of their arguments:

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

$\ln \left(\frac{x + 1}{\left(x - 1\right) \left(x - 1\right) \left(x + 1\right)}\right)$

Cancel the common factor $x + 1$:

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

Write the denominator as a square:

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

Because $1 = {1}^{2}$, we can write it this way:

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

Use the property $\ln \left({a}^{c}\right) = \left(c\right) \ln \left(a\right)$ to bring the -2 outside:

$- 2 \ln \left(x - 1\right)$