That's Not The Right Base

$\begin{aligned} \log_7 i & = \frac {i \pi} {\ln A} \\ \frac {\ln i} {\ln 7}&= \frac {i \pi} {\ln A} \\ \frac {\ln e^{i \frac \pi 2}} {\ln 7}&= \frac {i \pi} {\ln A} \\ \frac {i \pi } {2\ln 7}&= \frac {i \pi} {\ln A} \\ \ln A & = 2\ln 7 \\ \implies A&=\boxed {49}\end{aligned}$

In this problem we are getting $\Large{log_7 i = \frac{i \pi}{ln(A)} }$ . Now we will get

$log_7 i = log_e e^{i \pi} / log_e A$

Now it can be written as $log_7 i = log_A e^{i \pi}$

Now we know $e^{i \pi} = -1 [ cos \pi + i sin \pi ]$ By De Moivre's Theorem.

Putting this value we will get $log_7 i = log_A -1$

Now $i = 7^{log_A (-1)}$ .

Hence $i = (-1)^{log_A 7}$ .

We know $i = (-1)^{1/2}$

Or $log_A 7 = 1/2$ or $A^{1/2} = 7$ or $\large \boxed {A =49}$