.76 = 1 (1/2)^(t/5730)
log both sides:
log .76 = (t/5730) log .5
t/5730 = log .76/log .5 = .39592...
t = 2268.67 years
or
t = appr 2300 years, correct to the nearest 100 yrs.
An artifact was found and tested for its carbon-14 content. If 76% of the original carbon-14 was still present, what is its probable age (to the nearest 100 years)? (Carbon-14 has a half-life of 5,730 years.)
1 answer