Math  /  Algebra

QuestionPrehistoric cave paintings were discovered in a cave in France. The paint contained 19% of the original carbon-14. Use the exponential decay model for carbon-14, A-A000012 to estimate the age of the paintings The paintings are approximately years old. (Round to the nearest integer.)

Studdy Solution

STEP 1

1. The exponential decay model for carbon-14 is used, which is given by the formula N(t)=N0ekt N(t) = N_0 e^{-kt} .
2. N(t) N(t) is the amount of carbon-14 remaining at time t t .
3. N0 N_0 is the original amount of carbon-14.
4. k k is the decay constant for carbon-14.
5. The half-life of carbon-14 is approximately 5730 years.
6. The paint contains 19% of the original carbon-14.

STEP 2

1. Determine the decay constant k k .
2. Set up the exponential decay equation using the given percentage.
3. Solve for the age t t of the paintings.
4. Round the result to the nearest integer.

STEP 3

Determine the decay constant k k using the half-life of carbon-14.
The decay constant k k is related to the half-life T1/2 T_{1/2} by the formula:
k=ln(2)T1/2 k = \frac{\ln(2)}{T_{1/2}}
Substitute the half-life of carbon-14:
k=ln(2)5730 k = \frac{\ln(2)}{5730}

STEP 4

Set up the exponential decay equation using the given percentage.
Since the paint contains 19% of the original carbon-14, we have:
N(t)N0=0.19 \frac{N(t)}{N_0} = 0.19
Using the exponential decay model:
0.19=ekt 0.19 = e^{-kt}

STEP 5

Solve for the age t t of the paintings.
Take the natural logarithm of both sides:
ln(0.19)=kt \ln(0.19) = -kt
Substitute the expression for k k :
ln(0.19)=(ln(2)5730)t \ln(0.19) = -\left(\frac{\ln(2)}{5730}\right)t
Solve for t t :
t=ln(0.19)ln(2)5730 t = \frac{\ln(0.19)}{-\frac{\ln(2)}{5730}}
Calculate t t :
t1.66070.00012113788 t \approx \frac{-1.6607}{-0.000121} \approx 13788

STEP 6

Round the result to the nearest integer.
The estimated age of the paintings is approximately:
13788 \boxed{13788}

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