Answer
$= 1 + 2\log x + \frac{1}{3}\log(1-x) - \log7-2\log(x+1)$
Work Step by Step
$= \log[\frac{10x^{2}\sqrt[3] {1-x}}{7(x+1)^{2}}]$
$= \log(10x^{2}\sqrt[3] {1-x}) - \log(7(x+1)^{2})$
$= \log(10x^{2}(1-x)^{\frac{1}{3}}) - \log(7)-\log((x+1)^{2})$
$=\log(10x^{2})+\log(1-x)^{\frac{1}{3}} - \log7-2\log(x+1)$
$=\log(10) + \log x^{2} + \frac{1}{3}\log(1-x) - \log7-2\log(x+1)$
$= 1 + 2\log x + \frac{1}{3}\log(1-x) - \log7-2\log(x+1)$
