McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
1. Multiplying and Dividing Rational Expressions
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Exercise 70 Page 537

Use the exponential decay formula to determine the equation of decay for Carbon-14.

About 44 153 years ago.

Practice makes perfect

First, we will determine the equation of decay for Carbon-14. Then we will use it to answer the given question.

Equation of Decay for Carbon-14

We will use the exponential decay formula to determine the equation of decay for Carbon-14. y=ae^(- kt)Here, a is the initial value, t is time in years, and k is a constant representing the rate of continuous decay. To determine the equation of decay for Carbon-14, we have to find the value of k. We know that the half-life of Carbon-14 is 5760 years. It means that when t=5760 y is equal to the half of the initial value, y=0.5a.

y=ae^(- kt)
0.5a=ae^(- k(5760))
â–¼
Solve for k
0.5=e^(- k(5760))

ln(LHS)=ln(RHS)

ln0.5=lne^(- k(5760))

ln(e^a) = a

ln0.5=- k(5760)
ln0.5/- 5760=k
k=ln0.5/- 5760
k≈ 0.00012

We have that k is about 0.00012. We can write the equation of decay for Carbon-14. y=ae^(- 0.00012t)

How Long Ago Did The Person Die?

We know that the bones contain less than 0.5 % of the amount of Carbon-14 they would have contained when the person was alive. In other words, after t years the bones contain less than 0.005 of the initial amount a of Carbon-14. Let's substitute 0.005a for y into the equation of decay for Carbon-14. 0.005a=ae^(- 0.00012t) We want to find the value of t, so let's solve the above equation.

0.005a=ae^(- 0.00012t)
â–¼
Solve for t
0.005=e^(- 0.00012t)

ln(LHS)=ln(RHS)

ln0.005=lne^(- 0.00012t)

ln(e^a) = a

ln0.005=- 0.00012t
ln0.005/- 0.00012=t
t=ln0.005/- 0.00012
t≈ 44 153

The person died about 44 153 years ago.