Why doesn't this proof that $a \equiv b \pmod{m}$ for any integers $a, b, m$ work?

Question

As an example, lets take the fact that $10^{45} \equiv 6 \pmod{7}$: I'll use the fake proof to disprove this to show how it works.

First, we know that $10^{45} = (7^{\log_710})^{45} = 7^{45\log_710}$.

So we substitute in and get $7^{45\log_710} \equiv 6 \pmod{7}$.

We know that $a \equiv a'\pmod{c} \land b \equiv b'\pmod{c} \rightarrow a + b \equiv a' + b' \pmod{c}$, and so we can use this to conclude that:

$7^{45\log_710} \equiv 0^{45\log_710} \equiv 0 \pmod{7}$

But $0 \not\equiv 6 \pmod{7}$!

I feel like the answer is in the irrational exponent, but I'm not sure.

Why doesn't this false proof work?

Answer 1

The problem is that $\log_7 (x)$ doesn't make sense mod $7$, unless $x=0$. $\log_7 (x) = y$ iff $7^y=x$, but $7^y=0^y=0$, so this doesn't make sense. Thus $\log_7 (10)$ is meaningless in the integers mod $7$.