Answer
$2+3\displaystyle \log x+\frac{1}{3}\log(5-x)-\log 3-2\log(x+7)$
Work Step by Step
$\displaystyle \log[\frac{100x^{3}\sqrt[3]{5-x}}{3(x+7)^{2}}]=\quad $...apply the Quotient Rule: $\displaystyle \quad \log_{b}(\frac{M}{N})=\log_{b}\mathrm{M}-\log_{b}\mathrm{N}$
$=\log(100\cdot x^{3}\cdot\sqrt[3]{5-x})-\log[3\cdot(x+7)^{2}]$
$\quad $...apply the Product Rule: $\quad \log_{b}(MN)=\log_{b}\mathrm{M}+\log_{b}\mathrm{N}$
$=\log 100+\log x^{3}+\log\sqrt[3]{5-x}-[\log 3+\log(x+7)^{2}]$
... $\log 100=\log 10^{2}=2$
... $\sqrt[3]{5-x}=(5-x)^{1/3}$
$=2+\log x^{3}+\log(5-x)^{1/3}-\log 3-\log(x+7)^{2}$
$\quad $...apply the Power Rule: $\quad \log_{b}(M^{p})=p\cdot\log_{b}\mathrm{M}$
$=2+3\displaystyle \log x+\frac{1}{3}\log(5-x)-\log 3-2\log(x+7)$
