How do you condense 2(3lnx-ln(x+1)-ln(x-1))?

Apr 24, 2017

Depends how far you want to take things but as a single logarithm it becomes $\ln {\left(\frac{{x}^{3} \left(x - 1\right)}{x + 1}\right)}^{2}$

Explanation:

Multiples of logarithms become powers:
$2 \left(3 \ln \left(x\right) - \ln \left(x + 1\right) - \ln \left(x - 1\right)\right)$
$2 \left(\ln \left({x}^{3}\right) - \ln \left(x + 1\right) - \ln \left(x - 1\right)\right)$

Subtracting logarithms is equivalent to dividing their arguments:
$2 \left(\ln \left(\frac{{x}^{3}}{x + 1}\right) - \ln \left(x - 1\right)\right)$

Now divide again:
$2 \ln \left({x}^{3} / \left(\left(x + 1\right) \left(x - 1\right)\right)\right)$

Tidy this up to give:
$2 \ln \left(\frac{{x}^{3}}{{x}^{2} - 1}\right)$

You can apply the power law again:
$\ln \left(\frac{{x}^{6}}{{x}^{2} - 1} ^ 2\right)$

Apr 24, 2017

$\ln \left({x}^{6} / {\left({x}^{2} - 1\right)}^{2}\right)$

Explanation:

Expression $= 2 \left(3 \ln \left(x\right) - \ln \left(x + 1\right) - \ln \left(x - 1\right)\right)$

Here we will use three properties of logarithms:

(i): $a \ln x = \ln {x}^{a}$

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

(iii: $\ln a - \ln b \equiv \ln \left(\frac{a}{b}\right)$

First let's rewite the Expression as:

= $2 \left(3 \ln \left(x\right) - \left(\ln \left(x + 1\right) + \ln \left(x - 1\right)\right)\right)$

Using property (i):

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

Using properties (ii) and (iii):

= 2*lnx^3/(ln((x+1)(x-1))

= 2* lnx^3/(ln(x^2-1)

Using property (i) again:

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

$= \ln \left({x}^{6} / {\left({x}^{2} - 1\right)}^{2}\right)$