you could use
amount = initial(1/2)^(t/5730)
.88 = 1(1/2)^(t/5730)
t/5730ln.5 = ln .88
t = 5730ln.88/ln.5 = 1056.75 yrs.
or
let amount = initial (e^(kt))
when t = 5730 amount = .5
.5 = e^(5730k)
k = ln.5/5730 = -0.000120968
.88 = e^(-.000120968t)
t = ln.88/-.000120968
= 1056.75
An artifact was found and tested for its carbon-14 content. If 88% of the original carbon-14 was still present, what is its probable age (to the nearest 100 years)? Use that carbon-14 has a half-life of 5,730 years.
1 answer