Step 2
Since the half-life is 30 years, then [tex]y(30)=[/tex] ______ 5 mg.
Step 3
Thus, [tex]5=10 e^{30 k}[/tex], or [tex]e^{30 k}=[/tex] ______.
Therefore, [tex]k=[/tex] ______.
Answer (1)
The amount of the substance after 30 years is y ( 30 ) = 5 mg.
Substituting this into the equation y ( t ) = 10 e k t , we find e 30 k = 2 1 .
Taking the natural logarithm, we solve for k and find k = 30 l n ( 2 1 ) .
Therefore, k = 30 l n ( 0.5 ) ≈ − 0.0231 .
Explanation
Amount after half-life We are given that the half-life of the substance is 30 years. This means that after 30 years, half of the initial amount remains. Since the initial amount is 10 mg, after 30 years, the amount remaining is 5 mg. Therefore, y ( 30 ) = 5 mg.
Finding e^(30k) We are given the equation y ( t ) = 10 e k t . We know that y ( 30 ) = 5 , so we can substitute these values into the equation to get 5 = 10 e 30 k . Dividing both sides by 10, we get e 30 k = 10 5 = 2 1 .
Solving for k Now we need to find the value of k . We have e 30 k = 2 1 . To solve for k , we take the natural logarithm of both sides: ln ( e 30 k ) = ln ( 2 1 ) . Using the property of logarithms, we have 30 k = ln ( 2 1 ) . Dividing both sides by 30, we get k = 30 l n ( 2 1 ) . Since ln ( 2 1 ) = − ln ( 2 ) , we can write k = 30 − l n ( 2 ) . The approximate value of k is -0.0231.
Examples
Understanding exponential decay is crucial in various fields. For instance, in environmental science, it helps in predicting the decrease of pollutants in a lake over time. Imagine a lake initially contaminated with 100 units of a pollutant, and the pollutant's concentration decreases by half every year. Using the principles of exponential decay, we can model and predict how long it will take for the pollutant level to reach a safe threshold, say 1 unit, guiding remediation efforts and ensuring ecosystem health. The decay is modeled by the equation P ( t ) = P 0 e k t , where P ( t ) is the pollutant level at time t, P 0 is the initial pollutant level, and k is the decay constant.
