I am trying to prove the following claim:
Let $(G,*)$ be a cyclic group of size $m$ with generator $g$. Assume there exists some adversary $A'$ of size $T'=\frac{\left(T-O\left(\log m\right)\right)}{2}$, for some $T$, such that $$\mathbb{P}_{b\gets G}\left[A'(b)=\log_{g}(b)\right]>\frac{1}{2}.$$ Show that there exists an adversary $A$ of size $T$ such that $$\mathbb{P}_{b\gets G}\left[A(b)=\log_{g}(b)\right]>\frac{3}{4}.$$ Assume multiplication over $G$ requires $O(1)$ circuit.
How can I prove this claim? Intuitively, since $A$ is bigger than $A'$ in about $O(\log m)$, the construction should be based on dividing the problem into halfs recursively, as in binary search. However, I can't think of any clever way to do that...