smc

Since all four terms on the left are positive integers, from $\sqrt{\log a}$, we know that both $\log a$ has to be a perfect square and $a$ has to be a power of ten. The same applies to $b$ for the same reason. Setting $a$ and $b$ to $10^x$ and $10^y$, where $x$ and $y$ are the perfect squares, $ab=10^{x+y}$. By listing all the perfect squares up to $14^2$ (as $15^2$ is larger than the largest possible sum of $x$ and $y$ of $200$ from answer choice $\text{E}$), two of those perfect squares must add up to one of the possible sums of $x$ and $y$ given from the answer choices ($52$, $100$, $144$, $164$, or $200$). Only a few possible sums are seen: $16+36=52$, $36+64=100$, $64+100=164$, $100+100=200$, and $4+196=200$. By testing each of these (seeing whether $\sqrt{x}+\sqrt{y}+\frac{x}{2}+\frac{y}{2}=100$), only the pair $x=64$ and $y=100$ work. Therefore, $a$ and $b$ are $10^{64}$ and $10^{100}$, and our answer is $\boxed{10^{164}}$.