It's easy to figure out the answer which is 2. But I am trying to solve it in a different approach. My approach:
$2^x \equiv 2^4 + 2^4 \pmod 7
\Rightarrow x\log_2(2) \equiv 4\log_2(2) + 4\log_2(2) \pmod 7
\Rightarrow x \equiv 4 + 4 \mod 7
\Rightarrow x \equiv 8 \pmod 7
\Rightarrow x \equiv 1 \pmod 7 $
Here, the answer is 1(false) but it supposed to be 2. It works if I do the modular operation by 6.
PS: I read somewhere that it is related to Fermat's little theorem, but can't figure it out. Can someone give a detail explanation on this?
Best Answer
Fermat's little theorem says $2^6\equiv1\pmod7$. In fact, $2^3=8\equiv1\pmod7$.
$2^4+2^4=2\times2^4=2^5 $, so, because $2^3\equiv1\pmod7$, $2^{3n+5}\equiv2^4+2^4\pmod 7$ for all $n\in \mathbb Z$
$(n>-2$ to avoid negative exponents).
So if $2^x\equiv2^4+2^4\pmod7$, then $x$ could be $3m+2$
(integer $m>-1$ to avoid negative exponents).
On the other hand, $2^{3m}\equiv2^0\equiv1$ and $2^{3m+1}\equiv2^1\equiv2,$
so these are not congruent to $4=2^2\equiv2^5=2^4+2^4\pmod7$.