Please help
An anthropologist finds bone that her instruments measure it as 0.824% of the amount of Carbon-14 the bones would have contained when the person was alive. How long ago did the person die? The half life of carbon 14 is 5,730 years.

Question

Please help
An anthropologist finds bone that her instruments measure it as  0.824% of the amount of Carbon-14 the bones would have contained when the person was alive. How long ago did the person die? The half life of carbon 14 is 5,730 years.

phi · Answer

use 
$$ \text{ fraction left } = \left(\frac{1}{2}\right)^\frac{t}{5730} $$

kawaiiiimya · Answer

so (1/2) ^ 0.824/5730

phi · Answer

the 0.824% is how much is left of 100%
or as a fraction (divide by 100 to get rid of the % sign)
0.00824 is how much is left.  what we need to find is the time "t":
$$ 0.00824 = \left(\frac{1}{2}\right)^\frac{t}{5730}$$

kawaiiiimya · Answer

LN 0.00824=(T/5730) LN 1/2

phi · Answer

yes, that looks good

phi · Answer

people solve these types of problems with logarithms.
if you "take the log" of both sides
you would get
$$ \log(0.00824) = \log\left( \left(\frac{1}{2}\right)^\frac{t}{5730}\right) $$

kawaiiiimya · Answer

-4.798754935=(T/57300)(-0.693147181)

kawaiiiimya · Answer

6.923139943=(T/57300)

phi · Answer

ok, but it is 5730  (not 57300)
so 
T= 5730 * 6.923139943= 39,669.59 years
or about
T= 39,700 years