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# help!!

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A farmer in china discovers a mammal hide that contains 71% of its original amount of C - 14.

Find the age of the mammal hide to the nearest year.

Nov 16, 2018

#1
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71% =1 x (1/2)^(t/5730)
Solve for t :
0.71 = 2^(-t/5730)

0.71 = 71/100:
71/100 = 2^(-t/5730)

71/100 = 2^(-t/5730) is equivalent to 2^(-t/5730) = 71/100:
2^(-t/5730) = 71/100

Take reciprocals of both sides:
2^(t/5730) = 100/71

Take the logarithm base 2 of both sides:
t/5730 = log(100/71)/log(2)

Multiply both sides by 5730:

t = (5730 log(100/71))/log(2)=2831.24 =~2,831 Years

Nov 16, 2018

#1
+1

71% =1 x (1/2)^(t/5730)
Solve for t :
0.71 = 2^(-t/5730)

0.71 = 71/100:
71/100 = 2^(-t/5730)

71/100 = 2^(-t/5730) is equivalent to 2^(-t/5730) = 71/100:
2^(-t/5730) = 71/100

Take reciprocals of both sides:
2^(t/5730) = 100/71

Take the logarithm base 2 of both sides:
t/5730 = log(100/71)/log(2)

Multiply both sides by 5730:

t = (5730 log(100/71))/log(2)=2831.24 =~2,831 Years

Guest Nov 16, 2018
#2
+19325
+1

Fo =1/2 ^(t/5730)

.71 = 1/2^(t/5730)   take log of both sides

log(.71)/log(.5) = t/5730

t = 5730 * log (.71)/log(.5) =~~ 2831. years

Nov 16, 2018