The given equation is
[tex]\begin{gathered} y=log_bx \\ \text{where b > 1} \end{gathered}[/tex]
To know which of the options is the answer we substitute for x and y independently in the equation
For the graph, the values of the first option agrees with a value for b > 1
[tex]\begin{gathered} -3=1og_b\frac{1}{8} \\ b^{-3}=\frac{1}{8} \\ b^{-3}=2^{-3} \\ b\text{ =2} \end{gathered}[/tex]
Again
[tex]\begin{gathered} \text{if y=}1\text{ and x =}2 \\ 1=\log _b2 \\ b^1=2 \\ b\text{ =2} \end{gathered}[/tex]
So the answer is option A
