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Sagot :
Certainly! Let's solve the given problem step-by-step.
### Part (a)
We are given that the decomposition of the substance follows a continuous exponential decay model. The general formula for exponential decay is:
[tex]\[ y = A e^{kt} \][/tex]
where:
- [tex]\( y \)[/tex] is the amount of the substance at time [tex]\( t \)[/tex],
- [tex]\( A \)[/tex] is the initial amount of the substance,
- [tex]\( k \)[/tex] is the decay constant, and
- [tex]\( t \)[/tex] is time.
Given:
- [tex]\( A = 400 \)[/tex] kg (initial amount),
- After 7 hours, [tex]\( y = 260 \)[/tex] kg.
We need to determine the decay constant [tex]\( k \)[/tex]. Plugging the known values into the decay formula:
[tex]\[ 260 = 400 e^{7k} \][/tex]
To solve for [tex]\( k \)[/tex], first isolate the exponential term:
[tex]\[ \frac{260}{400} = e^{7k} \][/tex]
Simplify the fraction:
[tex]\[ \frac{13}{20} = e^{7k} \][/tex]
Now take the natural logarithm of both sides to solve for [tex]\( k \)[/tex]:
[tex]\[ \ln\left(\frac{13}{20}\right) = 7k \][/tex]
[tex]\[ k = \frac{1}{7} \ln\left(\frac{13}{20}\right) \][/tex]
Thus, the formula relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex] is:
[tex]\[ y = 400 e^{\left(\frac{1}{7} \ln\left(\frac{13}{20}\right) t\right)} \][/tex]
### Part (b)
Now we need to determine how much of the substance is left 13 hours after the beginning of the experiment.
We use the formula we found in part (a):
[tex]\[ y = 400 e^{\left(\frac{1}{7} \ln\left(\frac{13}{20}\right) \cdot 13\right)} \][/tex]
Let's denote the decay constant as [tex]\( k \)[/tex] again for simplicity. Given:
[tex]\[ k = \frac{1}{7} \ln\left(\frac{13}{20}\right) \][/tex]
Plug [tex]\( k \)[/tex] back into our formula with [tex]\( t = 13 \)[/tex]:
[tex]\[ y = 400 e^{13 \cdot \frac{1}{7} \ln\left(\frac{13}{20}\right)} \][/tex]
Simplify the exponent:
[tex]\[ y = 400 e^{\left(\frac{13}{7} \ln\left(\frac{13}{20}\right)\right)} \][/tex]
Using the property of logarithms and exponents [tex]\( a \ln(b) = \ln(b^a) \)[/tex]:
[tex]\[ y = 400 \left( e^{\ln\left(\left(\frac{13}{20}\right)^{\frac{13}{7}}\right)}\right) \][/tex]
Since [tex]\( e^{\ln(x)} = x \)[/tex]:
[tex]\[ y = 400 \left(\left(\frac{13}{20}\right)^{\frac{13}{7}}\right) \][/tex]
To find the exact value, calculate [tex]\( \left(\frac{13}{20}\right)^{\frac{13}{7}} \approx 0.45 \)[/tex]:
[tex]\[ y \approx 400 \times 0.45 \][/tex]
Finally, round the result to the nearest whole number:
[tex]\[ y \approx 180 \, \text{kg} \][/tex]
So, 13 hours after the beginning of the experiment, approximately 180 kg of the substance is left.
### Part (a)
We are given that the decomposition of the substance follows a continuous exponential decay model. The general formula for exponential decay is:
[tex]\[ y = A e^{kt} \][/tex]
where:
- [tex]\( y \)[/tex] is the amount of the substance at time [tex]\( t \)[/tex],
- [tex]\( A \)[/tex] is the initial amount of the substance,
- [tex]\( k \)[/tex] is the decay constant, and
- [tex]\( t \)[/tex] is time.
Given:
- [tex]\( A = 400 \)[/tex] kg (initial amount),
- After 7 hours, [tex]\( y = 260 \)[/tex] kg.
We need to determine the decay constant [tex]\( k \)[/tex]. Plugging the known values into the decay formula:
[tex]\[ 260 = 400 e^{7k} \][/tex]
To solve for [tex]\( k \)[/tex], first isolate the exponential term:
[tex]\[ \frac{260}{400} = e^{7k} \][/tex]
Simplify the fraction:
[tex]\[ \frac{13}{20} = e^{7k} \][/tex]
Now take the natural logarithm of both sides to solve for [tex]\( k \)[/tex]:
[tex]\[ \ln\left(\frac{13}{20}\right) = 7k \][/tex]
[tex]\[ k = \frac{1}{7} \ln\left(\frac{13}{20}\right) \][/tex]
Thus, the formula relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex] is:
[tex]\[ y = 400 e^{\left(\frac{1}{7} \ln\left(\frac{13}{20}\right) t\right)} \][/tex]
### Part (b)
Now we need to determine how much of the substance is left 13 hours after the beginning of the experiment.
We use the formula we found in part (a):
[tex]\[ y = 400 e^{\left(\frac{1}{7} \ln\left(\frac{13}{20}\right) \cdot 13\right)} \][/tex]
Let's denote the decay constant as [tex]\( k \)[/tex] again for simplicity. Given:
[tex]\[ k = \frac{1}{7} \ln\left(\frac{13}{20}\right) \][/tex]
Plug [tex]\( k \)[/tex] back into our formula with [tex]\( t = 13 \)[/tex]:
[tex]\[ y = 400 e^{13 \cdot \frac{1}{7} \ln\left(\frac{13}{20}\right)} \][/tex]
Simplify the exponent:
[tex]\[ y = 400 e^{\left(\frac{13}{7} \ln\left(\frac{13}{20}\right)\right)} \][/tex]
Using the property of logarithms and exponents [tex]\( a \ln(b) = \ln(b^a) \)[/tex]:
[tex]\[ y = 400 \left( e^{\ln\left(\left(\frac{13}{20}\right)^{\frac{13}{7}}\right)}\right) \][/tex]
Since [tex]\( e^{\ln(x)} = x \)[/tex]:
[tex]\[ y = 400 \left(\left(\frac{13}{20}\right)^{\frac{13}{7}}\right) \][/tex]
To find the exact value, calculate [tex]\( \left(\frac{13}{20}\right)^{\frac{13}{7}} \approx 0.45 \)[/tex]:
[tex]\[ y \approx 400 \times 0.45 \][/tex]
Finally, round the result to the nearest whole number:
[tex]\[ y \approx 180 \, \text{kg} \][/tex]
So, 13 hours after the beginning of the experiment, approximately 180 kg of the substance is left.
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