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Sagot :
To estimate the age of the prehistoric cave paintings in France, we can use the exponential decay model for carbon-14 given by the equation:
[tex]\[ A = A_0 e^{-0.000121 t} \][/tex]
Where:
- [tex]\( A \)[/tex] is the final amount of carbon-14, which in this case is 5% of the original amount ([tex]\( A_0 \)[/tex]). So, [tex]\( A = 0.05 A_0 \)[/tex].
- [tex]\( A_0 \)[/tex] is the initial amount of carbon-14.
- [tex]\( 0.000121 \)[/tex] is the decay constant for carbon-14.
- [tex]\( t \)[/tex] is the time elapsed, which we need to find.
First, we need to solve for [tex]\( t \)[/tex]. Starting with the decay formula:
[tex]\[ 0.05 A_0 = A_0 e^{-0.000121 t} \][/tex]
Next, we divide both sides of the equation by [tex]\( A_0 \)[/tex]:
[tex]\[ 0.05 = e^{-0.000121 t} \][/tex]
To solve for [tex]\( t \)[/tex], we take the natural logarithm (ln) of both sides:
[tex]\[ \ln(0.05) = \ln(e^{-0.000121 t}) \][/tex]
Using the property of logarithms that [tex]\( \ln(e^x) = x \)[/tex], we get:
[tex]\[ \ln(0.05) = -0.000121 t \][/tex]
Now, solving for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(0.05)}{-0.000121} \][/tex]
Using the value of [tex]\( \ln(0.05) \approx -2.9957 \)[/tex], we calculate:
[tex]\[ t = \frac{-2.9957}{-0.000121} \][/tex]
This simplifies to:
[tex]\[ t \approx 24758.12 \][/tex]
Now, rounding to the nearest integer:
[tex]\[ t \approx 24758 \][/tex]
Therefore, the age of the prehistoric cave paintings is approximately [tex]\( 24758 \)[/tex] years old.
[tex]\[ A = A_0 e^{-0.000121 t} \][/tex]
Where:
- [tex]\( A \)[/tex] is the final amount of carbon-14, which in this case is 5% of the original amount ([tex]\( A_0 \)[/tex]). So, [tex]\( A = 0.05 A_0 \)[/tex].
- [tex]\( A_0 \)[/tex] is the initial amount of carbon-14.
- [tex]\( 0.000121 \)[/tex] is the decay constant for carbon-14.
- [tex]\( t \)[/tex] is the time elapsed, which we need to find.
First, we need to solve for [tex]\( t \)[/tex]. Starting with the decay formula:
[tex]\[ 0.05 A_0 = A_0 e^{-0.000121 t} \][/tex]
Next, we divide both sides of the equation by [tex]\( A_0 \)[/tex]:
[tex]\[ 0.05 = e^{-0.000121 t} \][/tex]
To solve for [tex]\( t \)[/tex], we take the natural logarithm (ln) of both sides:
[tex]\[ \ln(0.05) = \ln(e^{-0.000121 t}) \][/tex]
Using the property of logarithms that [tex]\( \ln(e^x) = x \)[/tex], we get:
[tex]\[ \ln(0.05) = -0.000121 t \][/tex]
Now, solving for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(0.05)}{-0.000121} \][/tex]
Using the value of [tex]\( \ln(0.05) \approx -2.9957 \)[/tex], we calculate:
[tex]\[ t = \frac{-2.9957}{-0.000121} \][/tex]
This simplifies to:
[tex]\[ t \approx 24758.12 \][/tex]
Now, rounding to the nearest integer:
[tex]\[ t \approx 24758 \][/tex]
Therefore, the age of the prehistoric cave paintings is approximately [tex]\( 24758 \)[/tex] years old.
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