Suppose $2^a \equiv 2^b\pmod{101}$. Is $a \equiv b \pmod {100}$ always true?
The first thing that came in my mind was Fermat's Little Theorem. WLOG $a\ge b$. Since $(101,2)=1$, dividing both sides by $2^b$ gives $$2^{a-b}\equiv 1\pmod {101}$$ Also, $$2^{100}\equiv 1\pmod {101} $$ by Fermat's Little Theorem.
How should I continue?
By repeatedly squaring and reducing modulo $101$ we obtain $$\eqalign{ 2^2&\equiv4\cr 2^4&\equiv16\cr 2^5&\equiv32\cr 2^{10}&\equiv1024\equiv14\cr 2^{20}&\equiv196\equiv-6\cr 2^{40}&\equiv36\cr 2^{50}&\equiv504\equiv-1\ .\cr}$$ Thus $$2^{20}\not\equiv1\quad\hbox{and}\quad 2^{50}\not\equiv1\ ;$$ this is sufficient to show that $2^k$ can never be $1$ modulo $101$ for $1\le k<100$, and so in your problem, $a-b$ must be a multiple of $100$.
For the theory behind this, see my answer here.