<s>[INST] Solving a congruence of the form $a^x = b \pmod m$ without indices or primitive roots Consider $9^x\equiv 7 \mod 19$. So $9^x\equiv 26 \equiv 45$, $9^{x-1} \equiv 5 \equiv 24 \equiv 43 \equiv 62 \equiv 81$, so $x=3$, and $19 \mid 722$.
But what I really want to solve is $12^x\equiv 17 \mod 25$. Using the same method, $$2^{2x}\cdot3^x \equiv 17 \equiv 42,$$ $$ 2^{2x- 1}3^{x-1}\equiv 7 \equiv 32,$$ $$2^{2x-6}3^{x-1}\equiv 1\equiv 26, $$ $$ 2^{2x-7}3^{x-1}\equiv 13 \equiv38,$$ $$ 2^{2x-8}3^{x-1} \equiv 19\equiv 44, $$ $$2^{2x-10}3^{x-1}\equiv 11 \equiv 36 \equiv (2^2)(3^2),$$ so $2x-10=2$, hence $x=6$, and $x-1=2$, so $x=3$. 
Why doesn't this work? Is it because $12$ is not the power of a prime and $9$ is? Any help is appreciated! 
 [/INST] It is probably quickiest & easiest to calculate the powers of $12$ modulo $25$ ..
\begin{eqnarray*}
12,19,3,11,7,9,8,21,2,24 \\
13,6,22,14,18,16,\color{red}{17} \cdots
\end{eqnarray*}
So
\begin{eqnarray*}
12^{17}=17 \pmod{25}.
\end{eqnarray*}
</s>