Can someone check my chem calculation answer?

Question

A mammoth skeleton has a carbon-14 content of 3.125% of that found in living organisms. When did the mammoth live?
_____ years ago

i got 28650

Answer 1

C14 has a half life of 5730 years.
% left = 1/2 (t/λ) where t is the time elapsed, and λ is the half life.
1/2 (t/5730) = 3.125%

Use the logarithm function to drop the exponent:
t/5730 log 1/2 = log .03125
t = (5730 * log .03125) / log 1/2)
t = -8624.509 / -0.3010 = 28650 years

Thus, you are correct!