hello i need help with this application question

Question

The charcoal from a tree killed in a volcanic eruption contained 66.7% of the carbon-14 found in living matter. How old is the tree, to the nearest year? Use 5700 years for the half-life of carbon-14.

Answer 1

Let A(t) = amount of Carbon 14 remaining t years after the tree died and let A₀ = amount of Carbon 14 when the tree died.

A(t) = A₀e^kt, for some constant k.

A(5700) = 0.5A₀ So, 0.5A₀ = A₀e^5700k.

Therefore, 0.5 = e^5700k. So, 5700k = ln(0.5). We have k ≈ -0.0001216.

A(t) = A₀e^-0.0001216t

When A(t) = 0.667A₀, we get 0.667 =e^-0.0001216t

So, . -0.0001216t = ln(0.667). Therefore, t = ln(0.667) / -0.0001216 = 3330.3

The tree died about 3330 years ago.

Answer 2

0.667a₀ = a₀(0.5)^t/5700

0.667 = (0.5)^t/5700

ln 0.667 = (t/5700) ln 0.5

-0.404965 ≈ (t/5700) ln 0.5

-2308.30182 ≈ ln 0.5 t

-2308.30182 ≈ -06931t

3330..1844 ≈ t