Ans: D
$\begin{array}{rcl}
y & = & ab^x \\
\log_5 y & = & \log_5(ab^x) \\
\log_5 y & = & x\log_5b + \log_5 a \\
\end{array}$
$\begin{array}{rcl}
y & = & ab^x \\
\log_5 y & = & \log_5(ab^x) \\
\log_5 y & = & x\log_5b + \log_5 a \\
\end{array}$
By considering the $y$ intercept of the graph, we have
$\begin{array}{rcl}
\log_5 a & = & 2 \\
a & = & 5^2 \\
a & = & 25
\end{array}$
