Precalculus Algebra

Question

Carbon-14 has a half-life of 5,730 years. A fossil is found that has 16% of the carbon-14 found in a living sample. How old is the fossil? (Round your answer to the nearest year.)

Answer 1

This is an exponential decay problem so we use the formula y=(A) e^kt where k is the growth factor and t is time and A is our initial value. By the definition of half life only half of the original amount remains so

0.5 = e^5730k take the natural log of both sides

ln(0.5) = ln( e^5730k)

ln(0.5) = 5730 k k= ln(.5)/5730= -.0001209

When there is only 16% left we have the equation

.16= e^.-0001209t Again take the natural log of both sides

ln (0.16) = ln e^-.0001209t.

ln (0.16) = - .0001209t

t= ln(0.16)/-.0001209 = 15,158 years