The carbon-14 content of a wooden harpoon handle found in an Inuit archaeological site was found to be...?

Carbon-14.....

Radioactive decay is a first-order process and described by this equation:
A = Ao e^(-kt) .......... where A are the activities or "amounts", k is the decay constant, t is time
The decay constant is related to the half-life by:
k = ln 2 / t½
k = 0.693 / 5730 yr = 1.21x10^-4 yr⁻¹

Since the unknown is the elapsed time, t, rearrange the equation as:
ln A = -kt + ln Ao
ln (A/Ao) = -kt
t = ln(A/Ao) / -k
t = ln(0.619/1) / -1.21x10^-4yr⁻¹
t = 3970 yr .............................................. 3965 yr rounded to three significant digits

B

Let n be the number of half-life.

(1/2)ⁿ = 61.9%
log(0.5)ⁿ = log(0.619)
n log(0.5) = log(0.619)
n = log(0.619) / log(0.5) = 0.692

Age = (5730 years) × 0.692 = 3965 years

The answer: B) 3960 years (the closest answer)

One half life is the time for one half of the radioactive element to decay. Let n be the number of half lifes. ½ ^n represents the fraction of the carbon-14 that has not decayed. The next step is to convert the percent to a decimal fraction. Then set these two equal to each other and solve for n

½ ^n = 0.619

Let’s take the log of both sides of this equation.
n * log ½ = log 0.619
n = log 0.619 ÷ log ½

This is approximately 0.69. Multiply the half life of carbon-14 by this number.

T = 5,730 *(log 0.619 ÷ log ½)

This is approximately 3965 years. B is the closest to this answer.