Respuesta :
Answer:
The wood was burned 4257 years ago.
Step-by-step explanation:
The amount of C-14 present can be modeled by the following exponential function.
[tex]C(t) = C_{0}e^{rt}[/tex]
In which C is the amount remaining, [tex]C_{0}[/tex] is the initial amount, r is the rate that the amount decrases, and t is the time in years.
The C – 14 has a half-life of 5700 years.
This means that
[tex]C(5700) = 0.5C_{0}[/tex]
How long ago was the wood burned?
To solve this equation for t when [tex]C(t) = 0.6C_{0}[/tex], the first step is finding the rate that the amount decreases. For this, we apply the half-life information in the equation.
First step: Find the rate that the amount of C-14 decreases.
[tex]C(t) = C_{0}e^{rt}[/tex]
[tex]0.5C_{0} = C_{0}e^{5700r}[/tex]
[tex]e^{5700r} = 0.5[/tex]
The next step here is applying ln, since ln and e are inverse functions.
[tex]\ln{e^{5700r}} = \ln{0.5}[/tex]
[tex]5700r = -0.6931[/tex]
[tex]r = \frac{-0.6931}{5700}[/tex]
[tex]r = -0.00012[/tex]
Final step: Find the time
We have that [tex]C(t) = 0.6C_{0}[/tex], [tex]r = -0.00012[/tex].
So:
[tex]C(t) = C_{0}e^{rt}[/tex]
[tex]0.6C_{0} = C_{0}e^{-0.00012t}[/tex]
[tex]e^{-0.00012t} = 0.6[/tex]
Applying the ln in both sides of the equality.
[tex]\ln{e^{-0.00012t}} = \ln{0.6}[/tex]
[tex]-0.00012t = -0.5108[/tex]
[tex]t = \frac{0.5108}{0.00012}[/tex]
[tex]t = 4257[/tex]
The wood was burned 4257 years ago.
