Find the remainder ($x$) $a\equiv x\pmod{73}$ where
$a^{100}\equiv 2\pmod {73}$
$a^{101}\equiv 69\pmod{73}$
$a^{100}\equiv 2\pmod {73}\Rightarrow \phi(73)=72\Rightarrow 100x+72y=(100,72)=4$
This Diophantine equation doesn't have a solution.
Is it necessary to change the base (modulo)?
We know $$a^{101}\equiv 69\pmod{73},$$ and since $a^{101}\equiv a^{100}a\equiv 2a\pmod{73}$, we have the equation
$$2a\equiv 69\pmod{73}$$
This isn't too hard to just solve, but to make it a little easier, we can note $$69\equiv-4\pmod{73}$$ Hence $$2a\equiv-4\pmod{73}$$ and so finally $$a\equiv -2\pmod{73}$$
Since $-2\equiv 71\pmod{73}$, we have $71$ as the solution.