The terms are
1, 2, 4, 8, 16, 32........
The nth term appears to be 2^(n-1)
So....we need to solve this
2^(n - 1) = 5000 take the log of both sides
log 2^(n - 1) = log 5000 and we can write
(n - 1) log2 = log 5000
n - 1 = log 5000 / log 2
n = log 5000 / log 2 + 1 ≈ 13.28
So....the 14th term will exceed 5000