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.
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
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
+36916 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
