Congruences with exponents

Question

I'd like a clarification for this exercise.

$$ 7^x \equiv 19\mod30$$

He says: ''the $19$ has to be written like a power of $7$ otherwise it wouldn't be solvable.''

So he writes

$$ 7^x \equiv 7^2\mod30$$

First question: Why do I have to write $19$ as a power of $7$? And why $7^2$ and not $7^5$ ?

Then he proceeds by writing

\begin{align} 7^x-2 &\equiv 1\mod30\\ y&=x-2 \\ 7^y &\equiv 1\mod30\\ x-2 &\equiv 0\mod30\\ y &\equiv 0\mod4\end{align}

What is the reasoning in general about this procedure? What am I looking for when solving this type of exercise?

Answer 1

As $30\equiv2\cdot3\cdot5$

$$7^x\equiv19\pmod{30}\implies7^x\equiv19\pmod2\iff1^x\equiv1$$ which holds true for all integer $x$

Similarly for $\pmod3$ the condition holds true for all integer $x$

$$7^x\equiv19\pmod{30}\implies7^x\equiv19\pmod5$$

$$\iff2^x\equiv4\equiv2^2\implies x\equiv2\pmod{\phi(5)}$$

Answer 2

First of all, since $7^2 = 49 = 19 ~\mbox{mod}~ 30$ (The remainder when dividing 49 by 30), so we can replace 19 by $7^2$. So that $$7^x \equiv 7^2~\mbox{mod}~ 30 $$ dividing by $7^2$, we get $$7^{(x-2)} \equiv 1~\mbox{mod}~ 30 $$ Let $y = x- 2$, then $$7^y \equiv 1~\mbox{mod}~ 30 $$ So that $y = x-2 \equiv 0 ~\mbox{mod}~ 30$ (As the way of dealing with normal power).