Question

The amount of carbon-14 present in animal bones after t years is given by A(t)=A_0e^-0.00012t. A sample of fossil had 27% of the carbon 14 of a contemporary living sample. Estimate the age of the sample.

Answer 1

`t\text{ }\cong\text{ 10908 years}`

Step-by-step explanation

We have that the amount of Carbon-14 present in animal bones after t years is given by:

`A(t)=A_0e^(-0.00012t)`

Where A(t) = amount after t years

A0 = original amount of sample

t = number of years

Now, we have that the amount of a sample of fossil had 27% of carbon remaining in it.

This means that:

A(t) = 27% of A0

=> A(t) = 0.27A0

Putting that in the equation:

`\begin{gathered} \Rightarrow0.27A_0=A_0e^(-0.00012t) \\ \Rightarrow\text{ }(0.27A_0)/(A_0)=e^(-0.00012t) \\ 0.27=e^(-0.00012t) \\ \text{Now, find the natural logarithm of both sides:} \\ \Rightarrow\text{ ln(0.27) = -0.00012t} \\ \Rightarrow\text{ -1.309 = -0.00012t} \\ Divide\text{ through by -0.00012:} \\ t\text{ = }(-1.309)/(-0.00012) \\ t\text{ }\cong\text{ 10908 years} \end{gathered}`

The age of the sample is 10,908 years.