algebra precalculus - A geometric sequence has first term $256$ and ratio $0.75$. Find the smallest $n$ for which the sum of the first $n$ terms exceeds $1000$.

A geometric sequence whose first term = 256 and whose common ratio is 0.75. Find the smallest number of n for which the sum of the first n terms of the sequence exceeds 1000.

My turn:

$$S_n = \frac{256(1-(0.75)^n)}{1-0.75} > 1000$$
$$n \log{0.75} < \log{\frac{3}{128}}$$
$$n < 13.04$$ then
$$n = 13$$

What is wrong with my solution because 13 does not satisfy the requires but 14 does ?