SOLUTION: I have used this sight before but it has been awhile. I having trouble with this carbon-14 dating question. Can anyone help me. We are using 5715 has a half life of carbon-14.

Question 64487This question is from textbook college algebra
: I have used this sight before but it has been awhile. I having trouble with this carbon-14 dating question. Can anyone help me.
We are using 5715 has a half life of carbon-14.
A mummy was found with only 24.6% of carbon-14. How long ago did the person die, to the nearest year.
Thanks This question is from textbook college algebra

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
We are using 5715 has a half life of carbon-14.
A mummy was found with only 24.6% of carbon-14. How long ago did the person die, to the nearest year.
:
The half-life decay equation that I am familiar with:
A = Ao*2^(-t/h)
Where:
A = resulting amt
Ao = Original amt
t = time in yrs (5715 for carbon 14)
h = half-life in yrs
:
In the this problem result can be given in a decimal of .246, so Ao = 1.00
:
So we have:
2^(-t/5715) = .246
Using natural logs equiv of exponents we have:
(-t/5715)*ln(2) = ln(.246)
-t/5715(.693147) = -1.40242
:
Get rid of the denominator and the negatives, mult equation by -5715
.693147t = -1.40242*-5715
.693147t = + 8014.85
t = 8014.85/.693147
t = 11,563 years
:
you can check this on a good calc; Enter 2^(-11563/5715) = .2459996
:
make sense to you? any questions?

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