Answer
$1+2\displaystyle \log x+\frac{1}{3}\log(1-x)-\log 7-2\log(x+1)$
Work Step by Step
$\displaystyle \log[\frac{10x^{2}\sqrt[3]{1-x}}{7(x+1)^{2}}]=\quad $...apply the Quotient Rule: $\displaystyle \quad \log_{b}(\frac{M}{N})=\log_{b}\mathrm{M}-\log_{b}\mathrm{N}$
$=\log(10\cdot x^{2}\cdot\sqrt[3]{1-x})-\log[7\cdot(x+1)^{2}]$
$\quad $...apply the Product Rule: $\quad \log_{b}(MN)=\log_{b}\mathrm{M}+\log_{b}\mathrm{N}$
$=\log 10+\log x^{2}+\log\sqrt[3]{1-x}-[\log 7+\log(x+1)^{2}]$
$=\log 10+\log x^{2}+\log\sqrt[3]{1-x}-\log 7-\log(x+1)^{2}$
... $\log 10$=1
... $\sqrt[3]{1-x}=(1-x)^{1/3}$
$=1+\log x^{2}+\log(1-x)^{1/3}-\log 7-\log(x+1)^{2}$
$\quad $...apply the Power Rule: $\quad \log_{b}(M^{p})=p\cdot\log_{b}\mathrm{M}$
$=1+2\displaystyle \log x+\frac{1}{3}\log(1-x)-\log 7-2\log(x+1)$
