Ans: B
According to the figure, we have $0 < b < 1$ and
According to the figure, we have $0 < b < 1$ and
$\begin{array}{rcl}
3 & = & ab^0 \\
a & = & 3
\end{array}$
Hence, we have
$\begin{array}{rcl}
y & = & 3b^x \\
\log_7 y & = & \log_7 (3b^x) \\
& = & \log_7 3 + \log_7 b^x \\
& = & x\log_7 b +\log_7 3
\end{array}$
Note that $\log_7 b$ and $\log_7 3$ are the slope and the $y$-intercept respectively. Since $0 < b < 1$, then $\log_7 b<0$. Hence the slope is negative. And since $\log_7 3$ is positive, then the $y$-intercept is positive. Hence, the answer is B.
