Answer to Question #346792 in Differential Equations for daniel

Question

determine the approximate age of a piece of burned wood, if it was found that 85.5% of the C-14 found in living trees of the same type had decayed.

Answer 1

$$N=N_0(2)^{-t/\tau}$$

The half-life of radiocarbon (14C) is $\tau=5730$ years.

$$(1-0.855)N_0=N_0(2)^{-t/5730}$$

$$(2)^{-t/5730}=0.145$$

$$t=-5730(\dfrac{\ln0.145}{\ln2})$$

$$t=15963\ years$$