Solve $2^x=13 \bmod 3^4$

Question

Solve $2^x=13\bmod 3^4$

I know $\log13=30\bmod 3^4$ and $\log16=15 \bmod 3^4 $

I've tried subbing $\log13/\log16$ for $2$ but I am not sure what to do next.

Answer 1

In this case you're looking for $\log_2{13} \pmod{3^4}$, and you can write $$\log_{2}{13}=\frac{\log 13}{\log 2}\pmod{3^4}$$ Moreover, we know $\log 16=\log 2^4 = 4\log 2$. So $\log 2 = \frac14\log 16$. Substituting this yields $$\log_2{13}=\frac{\log 13}{\frac14\log 16}=4\frac{\log 13}{\log 16}\pmod{3^4}$$

Answer 2

$$2^x=13\bmod 3^4\tag{1}$$

means that $2^x=3^4 k+13=81k+13$ for some $k\in\Bbb Z$.

For $81k+13$ to equal $2^x$, for some $k,x\in\Bbb Z$, $k$ must be odd ($81k+13$ must be even).

For $k=1$, it exists no $x\in\Bbb Z$ such that $81+13=2^x$.

But for $k=3$, you get $81\cdot 3+13=256=2^8=2^x$.

Hence, $x=8$ is one (integer) solution of (1).