How do you condense 4log(x)-2log((x^2)+1)+2log(x-1) ?

Apr 8, 2016

$\log \left(\frac{{x}^{4} {\left(x - 1\right)}^{2}}{{x}^{2} + 1} ^ 2\right) = \log \left(\frac{{x}^{6} - 2 {x}^{5} + {x}^{4}}{{x}^{4} + 2 {x}^{2} + 1}\right)$

Explanation:

First, get rid of all the coefficients for the logarithms

$4 \log x = \log {x}^{4}$
$- 2 \log \left({x}^{2} + 1\right) = \log {\left({x}^{2} + 1\right)}^{- 2}$
$2 \log \left(x - 1\right) = \log {\left(x - 1\right)}^{2}$

Now you can rewrite the equation above as

$4 \log x - 2 \log \left({x}^{2} + 1\right) + 2 \log \left(x - 1\right)$
$= \log {x}^{4} + \log {\left({x}^{2} + 1\right)}^{- 2} + \log {\left(x - 1\right)}^{2}$

Finally, knowing that adding together $\log$s is the same as having one $\log$ with all of the $x$ bits multiplied together inside,

$\log {x}^{4} + \log {\left({x}^{2} + 1\right)}^{- 2} + \log {\left(x - 1\right)}^{2}$
$= \log \left({x}^{4} {\left({x}^{2} + 1\right)}^{-} 2 {\left(x - 1\right)}^{2}\right)$
$= \log \left(\frac{{x}^{4} {\left(x - 1\right)}^{2}}{{x}^{2} + 1} ^ 2\right)$

You can expand this out if necessary, to

$\log \left(\frac{{x}^{6} - 2 {x}^{5} + {x}^{4}}{{x}^{4} + 2 {x}^{2} + 1}\right)$