Carbon-14 has a half-life of 5,730 years. A fossil is found that has 24% of the carbon-14 found in a living sample. How old is the fossil? (Round your answer to the nearest year.)

Answer 1

decay is exponential

A(t)=A₀*b^t ,,,,b is base, A_o amount at time=0

given 1/2=b⁵⁷³⁰

ln(1/2)=5730*ln(b)

b=e^ln(1/2)/5730=(1/2)^(1/5730)=.9988 approx.

given .24=A(t)/A(0)=b^t

t=ln(.24)/ln(b)=11,851 years approx.

Answer 2

Hey Myranda,

Hopefully this will help:

Step one: Plug in the numbers to the decay formula.

1. *** 0.24 = (0.5)^{t/5730 ***}

^{Step two: Convert the numbers and apply the arithmetic.}

1. 0.24 = (t/5730) in 0.5

Step three: Finish off.

1. 2.05889 ≈ t/5730
2. 11,797 ≈ t

Hopefully this helps.

Please let me know if you have any other questions. Thank you!