Answer
$1+\displaystyle \frac{1}{2}\log x $
Work Step by Step
... apply: $\sqrt{ab}=(ab)^{1/2}$
$\log\sqrt{100x}=\log(100x)^{1/2}$
$\quad $...apply the Power Rule: $\quad \log_{b}(M^{p})=p\cdot\log_{b}\mathrm{M}$
$=\displaystyle \frac{1}{2}\log(100x)$
$\qquad $...apply the Product Rule: $\qquad \log_{b}(MN)=\log_{b}\mathrm{M}+\log_{b}\mathrm{N}$
$=\displaystyle \frac{1}{2}(\log 100+\log x)$9 is
... log (without base) is $\log_{10}$
... $100=10^{2}$, and $\log_{b}b^{x}=x $
... so, $\log 100=\log_{10}10^{2}=2$
$=\displaystyle \frac{1}{2}(2+\log x)\qquad $... distribute
= $1+\displaystyle \frac{1}{2}\log x $
