Diffie-Hellman Challenge

Find the value of x, 1440<=x<M-1, which minimizes the value of y=(3^x % M), where % is the "modulo" or "take the remainder" operator, and M=2^2281-1. M is the 17th Mersenne prime. 3 is a primitive root of M. The value 1440 makes 3^x loop around and the modulo operation take effect, i.e., 3^1440>M. This could be solved once and for all by solving the discrete logarithm problem 3^x == 2 (mod M). If this challenge gets solved, or is too easy, then do the same thing with 3^x % M21701, where M21701=2^21701-1 is also a Mersenne prime. By some analysis I don't fully remember, I have reason to believe that 3 is probably a primitive root of M21701. (I think the reasoning was that if 3 is not primitive, it gets eliminated from consideration by some small factor of M-1, and the small factors of M21701 were tested.) And if that's still too easy, try the modulus P=5359*2^5054502+1, which is a prime number found by the Seventeen or Bust project. I don't know the least primitive root of P, but it should be pretty easy since P-1 factors trivially.

Update: Pohlig-Hellman makes this a bad idea.