Answer
$1+\frac{1}{2}log_{10}x$
Work Step by Step
We know that $log(x)$ is a common logarithm with an understood base of 10. Therefore, $log\sqrt (100x)=log_{10}\sqrt (100x)$.
Based on the product rule of logarithms, we know that $log_{b}(MN)=log_{b}M+log_{b}N$ (for $M\gt0$ and $N\gt0$).
Therefore, $log_{10}\sqrt (100x)=log_{10}\sqrt100+log_{10}\sqrt x=log_{10}10+log_{10}x^{\frac{1}{2}}$.
According to the power rule of logarithms, we know that $log_{b}M^{p}=plog_{b}M$ (when $b$ and $M$ are positive real numbers, $b\ne1$, and $p$ is any real number).
Therefore, $log_{10}10+log_{10}x^{\frac{1}{2}}=log_{10}10+\frac{1}{2}log_{10}x$.
Furthermore, based on the definition of the logarithmic function, we know that $y=log_{b}x$ is equivalent to $b^{y}=x$ (for $x\gt0$ and $b\gt0$, $b\ne1$). Therefore, $log_{10}10=1$, because $10^{1}=10$.
So, $log_{10}10+\frac{1}{2}log_{10}x=1+\frac{1}{2}log_{10}x$.
