An animal skull still has 20% of the carbon-14 that was present when the animal died. The half life of carbon-14 is 5730 years. Find the approximate age of the skull.
I understand I will need to use either log or ln but I don’t know how to get to that point. I think the formula y=ab^t will become something like y=a(1+r)^t and I think the y in this case is going to be 5730=a(1+(-.20)^……maybe i am backwrds maybe is is y=a(1+(-.20)^5730? Please help!!
✅ Answers
? Best Answer
t/5730 = log0.2/log0.5
t = 13304.65 years
general formula:
f(t) = Ae^(kt)
where f(t) = content of carbon-14 at time t
t = year
half-life = 5730 years –>
0.5A = Ae^(5730k)
0.5 = e^(5730k)
k = ln(0.5) / 5730
An animal skull still has 20% of the carbon-14
0.2A = Ae^(t*ln(0.5) / 5730)
0.2 = e^[t*ln(0.5) / 5730]
t = ln(0.2) / [ln(0.5) / 5730] = 5730*ln(0.2)/ln(0.5) = 13305 years (rounded)
let x be the number of half-lives it takes to decay to 20% of its original C-14 content.
2^-x = 0.2
-x log 2 = -log 5
x = log 5 / log 2
therefore age of skull = 5730x
= 5730 log 5 / log 2
= 13305 years
.5 = x^5730
log .5 =log x * 5730
log x = log .5/5730 = .000052536
Because 10^2 = 100 is the same as log 100 = 2
10^-.000052536 = x
.999879039 = x
.2 = .999879039^x
log .2 = log .999879039 * x
log .2/log .999879039 = x
x = 13,304.65