Conversion to discrete logarithm

Question

Given that $13^2 \equiv 2^3 \times 3^2 \pmod{97} $. How to derive to the form of discrete logarithm

$$2 \equiv 3 \log_{13}{2} + 2\log_{13}{3} \pmod{96}?$$

I cannot figure this out. The only thing I know is the following definition

$$a \equiv g^\mu \pmod{n} \Rightarrow \mu \equiv \log_{g}{a} \pmod{n}.$$

Answer 1

We know if $g$ is a primitive root $\pmod n,$

$a\equiv g^\mu\pmod n\implies\mu\equiv $ind$_ga\pmod{\phi(n)}$

So, here we need to establish that $13$ is a primitive root $\pmod{97}$

As $\phi(97)=96,$ it is sufficient to show that for all positive integer $r<96$ such that $r|96$ $$13^r\not\equiv1\pmod{97}$$

Answer 2

I think the result is true mod 97, not mod 96. Assuming mod 97 arithmetic, take logs to get $$ \log_{13} 13^2 = 2 \log_{13} 13 = 2 $$ and $$ \log_{13} \left(2^3 \times 3^2\right) = \log_{13} \left(2^3\right) + \log_{13} \left(3^2\right) = 3 \log_{13} 2 + 2 \log_{13} 3 $$