SOLUTION: This is a carbon decay question that I haven't encounted yet. Can anyone give me a hand here? Thanks! Question: Only 25% of the carbon-14 in a wooden bowl remains. How old i

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Question 62855This question is from textbook college algebra
: This is a carbon decay question that I haven't encounted yet. Can anyone give me a hand here? Thanks!
Question: Only 25% of the carbon-14 in a wooden bowl remains. How old is the bowl?
Thanks for your help!
This question is from textbook college algebra

Answer by ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!
This is a carbon decay question that I haven't encounted yet. Can anyone give me a hand here? Thanks!
:
Question: Only 25% of the carbon-14 in a wooden bowl remains. How old is the bowl?
:
The Half-life formula: A = Ao*.5^(t/h)
Where:
A = resulting amt; 25% here
Ao= Initial amt; 1 here
t = time (in years here)
h = half life, (carbon 14 half-life given as 5730 yrs)
:
.25 = 1 * .5^(t/5730)
:
Use the log equiv of exponent:
ln(.25) = (t/5730) * ln(.5)
-1.38629 = (t/5730) * -.693147
-1.38629 = -.693147t/5730
:
Get rid of denominator, mult equation by 5730
-1.3829*5730 = -.693147t
-7943.466689 = -.693147t
t = -7943.466689/-.693147
t = 11,460 yrs
:
Check:
A = 1*.5^(11460/5740); using a good calc
A = .25

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