We have the expression:
[tex]c^{(4d+1)}=7a-b[/tex]
We can apply logarithm to both sides. We would use it in order to get "4d+1". Then, we would apply logarithm with base c. This is beacuse of the definition of logarithm:
[tex]\log _c(x)=y\Leftrightarrow c^y=x[/tex]
If we apply this to our expression, we get:
[tex]\begin{gathered} c^{(4d+1)}=7a-b \\ \log _c(c^{(4d+1)})=\log _c(7a-b) \\ 4d+1=\log _c(7a-b) \end{gathered}[/tex]
If we rearrange both sides, we get the expression in Option B (we have to switch the sides):
[tex]\begin{gathered} 4d+1=\log _c(7a-b) \\ \log _c(7a-b)=4d+1 \end{gathered}[/tex]
Answer: Option B
