The function A=A0e^-0.01386x models the amount in pounds of a particular radioactive material stored in a concrete vault, where x is the number of years since the material was put into the vault. If 500 pounds of material are placed in the vault.
A) How much time will need to pass for only 47 pounds to remain?
B) How much material will be left after 10, 20, and 30 years?
A)
47 = 500 e^(-0.01386x)
take ln of both sides and use log rules
ln 47 = (-0.01386x)ln 500
-0.01386x = ln47/ln500
x =
continue
B)I will do the case of 20 years, you do the other two
A = 500 e^(-0.01386(20))
= 500 e^-.2772
= 500(.7579029..)
= appr 379 pounds
47 = 500 e^(-0.01386x)
e^(-0.01386x) = 47/500
-0.01386x = ln(47/500)
x = ln(0.094)/-0.01386
A) To find the time it will take for only 47 pounds to remain, we need to set up an equation using the given function.
We know that A = A0e^(-0.01386x), where A0 is the initial amount (500 pounds) and A is the amount remaining. We want to solve for x (the number of years).
Setting A = 47 pounds, we have:
47 = 500e^(-0.01386x)
To solve for x, we first divide both sides of the equation by 500:
47/500 = e^(-0.01386x)
Now, we can take the natural logarithm (ln) of both sides to isolate x:
ln(47/500) = ln(e^(-0.01386x))
Using the property of logarithms ln(e^x) = x, we simplify the equation to:
ln(47/500) = -0.01386x
Finally, we can solve for x by dividing both sides by -0.01386:
x = ln(47/500) / -0.01386
Using a calculator, we can evaluate the right side of the equation to find x ≈ 66.75 years.
Therefore, it will take approximately 66.75 years for only 47 pounds to remain.
B) To find the amount of material left after 10, 20, and 30 years, we can substitute these values into the given function.
For 10 years:
A = A0e^(-0.01386x)
A = 500e^(-0.01386 * 10)
A ≈ 500e^(-0.1386)
A ≈ 500 * 0.8706
A ≈ 435.3 pounds
After 10 years, approximately 435.3 pounds of material will be left.
For 20 years:
A = A0e^(-0.01386x)
A = 500e^(-0.01386 * 20)
A ≈ 500e^(-0.2772)
A ≈ 500 * 0.7579
A ≈ 378.95 pounds
After 20 years, approximately 378.95 pounds of material will be left.
For 30 years:
A = A0e^(-0.01386x)
A = 500e^(-0.01386 * 30)
A ≈ 500e^(-0.4158)
A ≈ 500 * 0.6581
A ≈ 329.05 pounds
After 30 years, approximately 329.05 pounds of material will be left.
To answer the question, let's break it down step by step.
A) How much time will need to pass for only 47 pounds to remain?
In the given function A = A0e^(-0.01386x), A represents the amount of radioactive material remaining after x years, A0 represents the initial amount of material, and e is Euler's number (approximately 2.71828).
We are given that A0 (initial amount of material) is 500 pounds, and we need to find the time (x) at which only 47 pounds will remain.
We can solve for x by rearranging the equation:
47 = 500e^(-0.01386x)
To isolate x, divide both sides by 500:
47/500 = e^(-0.01386x)
Next, take the natural logarithm (ln) of both sides to get rid of the exponential:
ln(47/500) = ln(e^(-0.01386x))
Applying the logarithmic property of ln(a^b) = b * ln(a), the equation becomes:
ln(47/500) = -0.01386x * ln(e)
Since ln(e) is equal to 1, we simplify further:
ln(47/500) = -0.01386x
Now, we isolate x by dividing both sides by -0.01386:
x = ln(47/500) / -0.01386
Using a calculator, compute the value of ln(47/500) and divide it by -0.01386 to find the value of x. This will give you the number of years needed for only 47 pounds to remain in the vault.
B) How much material will be left after 10, 20, and 30 years?
We can use the same function A = A0e^(-0.01386x) to find the amount of material remaining after a certain number of years.
For each given time, substitute the value of x into the equation and solve for A.
For 10 years:
A = 500e^(-0.01386 * 10)
For 20 years:
A = 500e^(-0.01386 * 20)
For 30 years:
A = 500e^(-0.01386 * 30)
Using a calculator, evaluate each of these equations to find the respective amount of material remaining after 10, 20, and 30 years.