Carbon-14 math problem

Question

The amount of​ carbon-14 present in animal bones after t years is given by ​P(t)=P0e−0.00012t. A bone has lost 11​% of its​ carbon-14. How old is the​ bone?

Answer 1

.11 = e^-.00012t

log.11 = -.00012t

t= log.11/-.00012 = about 7,989 years old

half life is when

.5P = Pe^-.00012t, P's cancel out leaving

.5 = e^-.00012t

take logs

log.5 = -.00012t

divide by -.00012 to get

t = log.5/-.00012 = about 2,509 years but half life of C14 is well known as 5730. half will decay into nitrogen 14

so after 5,730 years

mistake(s) somewhere above

formula is very accurate, even more accurate is Pt = Poe^-.000121t

but that doesn't explain the different answers. .11 is slightly over two half lives which leaves ..125, so age should be slightly over 11,460 years old, not as "young" as 7,989 yrs. Find the error(s) above and you win a free one way ticket to Gaza, the Ukraine, the Congo, NE Nigeria or a part of Afghanistan where the Taliban runs hog wild. Congrats. Baggage fees extra. Might be wise to take out some life insurance first, before boarding.

I just found the error. use ln not log, hit the ln key

t = ln.5/-.000121 = 5728.49 years old for half life

t = ln.5/-.00012 = 5776

for .11 left

t = ln.11/-.00012 = about 18,394 which is a little over the double 5730 = 11460 years old

in advanced math log = natural log base e, but on calculator log = common log base 10