Question

Suppose you find a piece of human bone that contains 48% of the amount of radioactive carbon-14 normally found in the bone of a living person. How old is the bone? (Half-life of carbon-14 is 5730 years)

Answer 1

use decay function
\[\large P_n = P_0 e^{-kt}\]
given the half life, you can find "k"
\[\large 0.5 = e^{-5730k}\]
\[k = \frac{\ln 0.5}{-5730} \]
now solve for "t"
\[\large 0.48 = e^{\frac{\ln 0.5}{5730} t}\]
\[t = \frac{5730 \ln 0.48}{\ln 0.5}\]

Answer 2

you can also do this without any \(e\)

Answer 3

\[\left(\frac{1}{2}\right)^{\frac{t}{5730}}=.48\] solve for \(t\)

Answer 4

change of base \[\frac{t}{5730}=\frac{\ln(.48)}{\ln(.5)}\]\[t=\frac{5730\ln(.48)}{\ln(.5)}\]

Answer 5

thanks to both of you soo much!! i actually understand it!