Question

A farmer in China discovers a mammalhide that contains 30% of its originalamount of C-14.N = Noe -ktNo inital amount of C-14 (at timeHt = 0)N= amount of C-14 at time tk= 0.0001t= time, in years

Answer 1

We will use the formula given

`\begin{gathered} N=N_0e^(-kt) \\ here\text{ N = 30\% of N}_0 \\ N=0.3N_0 \\ k=0.0001 \end{gathered}`

Plugging in we have

`\begin{gathered} N=N_0e^(-kt) \\ 0.3N_0=N_0e^(-0.0001t) \\ e^(-0.0001t)=(0.3N_0)/(N_0) \\ e^(-0.000t)=0.3 \end{gathered}`

Taking Ln of both sides we have

`\begin{gathered} ln(e^(-0.000t))=ln0.3 \\ -0.0001t=ln0.3 \\ t=(ln0.3)/(-0.0001) \\ t=12,039.728043 \end{gathered}`

Hence the answer is 12,040 years to the nearest year