This answer contains extra information, involving definitions, that you are not required to write in your answer.

S = {x | x ∈ {a, b, c}* and the length of x is a power of 2}.

Theorem. S is not regular.

Proof. A language S is regular if and only if there exists a finite-state machine M that solves S. Such a machine must answer yes on all strings that are in S and must answer no on all strings that are not in S.

The proof is by contradiction. Suppose that S is regular. Then there exists a finite-state machine M that solves S.

Run M on strings of a's whose lengths are powers of 2. (That is, run it on strings a¹, a², a⁴, a⁸, …)

For each of those strings, write down the state in which M ends. Stop when two different strings aⁱ and a^j have led M to stop in the same state, q, where i < j.

Since we only run M on strings whose length is a power of 2, there must be two integers u and v so that i = 2^u and j = 2^v.

Now run M on strings aⁱaⁱ and on a^jaⁱ. Since aⁱ and a^j take M to the same state q, aⁱaⁱ and a^jaⁱ must also take M to the same state, q'.

But aⁱaⁱ has a length that is a power of 2 (2^u+1). So q' must be an accepting state.

String a^jaⁱ cannot have a length that is a power of 2 since j + i = 2^v + 2^u, and the next power of 2 after 2^v is 2^v+1 > 2^v + 2^u. So q' must be a recting state.

That is a contradiction. q' cannot be both an accepting state and a rejecting state.

So the initial assumption (that S is regular) must be incorrect.

An abbreviated proof

The following shorter proof would be good.

Theorem. S is not regular.

Proof. By contradiction. Suppose that finite-state machine M solves S.

Run M on strings of a's whose lengths are powers of 2. There must be two different strings aⁱ and a^j, where i < j and i and j are powers of 2, that take M to the same state q. From state q, run M on a²ⁱ.

We find that strings a²ⁱ and a^j+i take M to the same state q'. String a²ⁱ is in S, so q' must be an accepting state.

Notice that j + i cannot be a power of 2, since the next power of 2 after j is 2j, which is larger than j + i. So a^j+i is not in S, and q' must be a rejecting state.

That is a contradiction.