.5 = e^-k(5730)
ln .5 = -.6931 = - 5730 k
k = 1.2097*10^-4
.45 = e^-1.2097*10^-4 t
ln .45 = -.7985 = -1.2097*10^-4 t
t = 6601 years
the amount of carbon-14 in a mammoth's bone is 45% of the amount found in a living organism. How long ago did the mammoth die? the half-life of carbon-14 is about 5,730 years.
2 answers
or, you can think of it like this: every half-life, 1/2 of what was there is gone. So, the amount left after t half-lives is
(1/2)^(t/5730)
So, we want t when
(1/2)^(t/5730) = .45
t/5730 = log(.45)/log(.5) = 1.152
That is, it takes 1.152 half-lives to reduce to 45%
So, t = 1.152 + 5730 = 6601
(1/2)^(t/5730)
So, we want t when
(1/2)^(t/5730) = .45
t/5730 = log(.45)/log(.5) = 1.152
That is, it takes 1.152 half-lives to reduce to 45%
So, t = 1.152 + 5730 = 6601