Answer
$\log ( \frac{x^4(x^2-1)}{\sqrt [3] {x^2+1}})$
Work Step by Step
$Combine$ $the$ $expression$:
$4$$\log x$ - $\frac{1}{3}$$\log (x^2+1)$ + $2$$\log (x-1)$
Apply the Third Law of Logarithms for $4$$\log x$, $\frac{1}{3}$$\log (x^2+1)$, and $2$$\log (x-1)$
$4$$\log x$ = $\log x^4$
$\frac{1}{3}$$\log (x^2+1)$ = $\log (x^2+1)^\frac{1}{3}$
$2$$\log (x-1)$ = $\log (x-1)^2$
$\log x^4$ - $\log (x^2+1)^\frac{1}{3}$ + $\log (x-1)^2$
$\log x^4$ - $\log \sqrt [3] {x^2+1}$ + $\log (x-1)^2$
Apply the Second Law of Logarithms for $\log x^4$ - $\log \sqrt [3] {x^2+1}$
$\log x^4$ - $\log \sqrt [3] {x^2+1}$ = $\log (\frac{x^4}{\sqrt [3] {x^2+1}})$
$\log (\frac{x^4}{\sqrt [3] {x^2+1}})$ + $\log(x-1)^2$
Apply the First Law of Logarithms
$\log (\frac{x^4}{\sqrt [3] {x^2+1}})$ + $\log(x-1)^2$ = $\log ( (\frac{x^4}{\sqrt [3] {x^2+1}}\times (x-1)^2)$
$\log ( (\frac{x^4}{\sqrt [3] {x^2+1}}\times \frac{(x-1)^2)}{1})$
$\log ( \frac{x^4(x^2-1)}{\sqrt [3] {x^2+1}})$
