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

Question

5860 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-08T11:19:32

$log((x^4(x -1)^2)/(x^2+1)^2) = log((x^6 - 2x^5 + x^4)/(x^4 + 2x^2 + 1))$

First, get rid of all the coefficients for the logarithms

$4logx = logx^4$
$-2log(x^2 + 1) = log(x^2 + 1)^(-2)$
$2log(x-1) = log(x-1)^2$

Now you can rewrite the equation above as

$4logx - 2log(x^2 + 1) + 2log(x-1)$
$= logx^4 + log(x^2 + 1)^(-2) + log(x-1)^2$

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

$logx^4 + log(x^2 + 1)^(-2) + log(x-1)^2$
$= log(x^4(x^2+1)^-2(x-1)^2)$
$= log((x^4(x-1)^2)/(x^2+1)^2)$

You can expand this out if necessary, to

$log((x^6 - 2x^5 + x^4)/(x^4 + 2x^2 + 1))$

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