A fossilized leaf is found to contain 10 micrograms of carbon-14 whereas a leaf of this type normally contains about 13.5 micrograms of carbon-14. Estimate the age of te fossilized leaf. (The half-life of carbon-14 is 5730 years)
+23254 FA = IA · (1/2)t/h
where FA = Final Amount = 10 IA = Initial Amount = 13.5
t = time (years) h = half-life (years) = 5730
---> 10 = 13.5 · (1/2)t/5730
---> 0.74074 = (1/2)t/5730 (divide both sides by 13.5)
---> log(0.74074) = log( (1/2)t/5730 ) (since the variable is in the exponent, find the log)
---> log(0.74074) = (t/5730)·log(1/2) (an exponent in a log comes out as a multiplier)
---> 5730·log(0.74074) = t·log(1/2) (multiply both sides by 5730)
---> 5730·log(0.74074)/log(1/2) = t (divide both sides by log(1/2)
---> t = <calculator time!>
+23254 FA = IA · (1/2)t/h
where FA = Final Amount = 10 IA = Initial Amount = 13.5
t = time (years) h = half-life (years) = 5730
---> 10 = 13.5 · (1/2)t/5730
---> 0.74074 = (1/2)t/5730 (divide both sides by 13.5)
---> log(0.74074) = log( (1/2)t/5730 ) (since the variable is in the exponent, find the log)
---> log(0.74074) = (t/5730)·log(1/2) (an exponent in a log comes out as a multiplier)
---> 5730·log(0.74074) = t·log(1/2) (multiply both sides by 5730)
---> 5730·log(0.74074)/log(1/2) = t (divide both sides by log(1/2)
---> t = <calculator time!>
