What Evil hints at is correct.  

If you have a generator element b of an additive group of order N, and you know it takes x repeated doubling operations (squaring operations in the context of a multiplicative group) on this element in order to reach q then you have effectively solved [multiplicative group notation]:

   q = b^2x

for x.

What we really want to do is crack the discrete logarithm, which means finding x in the context of:

   q = b^x

So the question becomes: if we can solve the first equation can we solve the second?  The answer is yes.  The reason is that the exponent of b is itself an element of it's own multiplicative group ℤ_N^× of which 2 is a generator since N is prime (at least in the context of Curve25519 or secp256k1).

So if you find x that satisfies:

   q = b^2x

then you can use your solution to solve the discrete log easy peasy lemon squeezy.:

   log_b(q) = 2^x mod N

Having at this point successfully solved the discrete log, you can happily collect your 30 BTC which will shortly be worthless since the security of bitcoin relies on the discrete logarithm assumption.