SOLUTION: Hi, I know that we are not supposed to put 2 questions in 1 however, they both go together so I felt as though it was best to put them both on one submission. The radioactive i

Question 1201494: Hi, I know that we are not supposed to put 2 questions in 1 however, they both go together so I felt as though it was best to put them both on one submission.
The radioactive isotope of potassium-42, which is vital in the diagnosis of brain tumors, has a half life of 12.36 hours.
Determine the exponential decay model that represents the mass of potassium-42.If 500-mg of potassium-42 was taken, how much milligrams of this isotope will remain after 48 hours?

Found 2 solutions by josgarithmetic, math_tutor2020:
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
To create equation using the given half life,

and you are asked y given x=48.

Not knowing how your computation is required to be made, you put into the Google search engine search or text field, this: 500*(0.5)^(48/12.36), and do what you need with the result.

Maybe, 34
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

x = number of hours that elapse
y = amount, in mg, of potassium-42 remaining
H = half-life of potassium-42 = 12.36 hours

Half-life template equation
y = a*(0.5)^(x/H)

We could use 1/2 in place of 0.5, but I find the 0.5 is easier to work with here.

Let's walk through a few scenarios for x.
If one half-life elapses, then x = H. The exponent x/H becomes H/H = 1, meaning we have one copy of 0.5 multiplied to the starting amount 'a'. Half of the substance remains.
If x = 2H, two half-lives elapse, and x/H becomes 2H/H = 2. We'll have 2 copies of 0.5 multiplied with the 'a'. Therefore, (0.5)^2 = 0.25 = 25% of the substance remains.
If x = 3H, three half-lives elapse, and x/H becomes 3H/H = 3. We'll have 3 copies of 0.5 multiplied with the 'a'. Therefore, (0.5)^3 = 0.125 = 12.5% of the substance remains.
And so on.

y = a*(0.5)^(x/H)
y = 500*(0.5)^(x/12.36)
This is one way to represent the exponential model, aka equation, aka function.

We have exponential decay because the base b = 0.5 fits the interval 0 < b < 1.

There are other formats to express this equation into.
Here are the steps to arrive at a second format.
y = 500*(0.5)^(x/12.36)
y = 500 * [ (0.5)^(1/12.36) ]^x
y = 500 * ( 0.94546361966231 )^x
The decimal value is approximate.
Round this value however your teacher instructs.

Here's yet another pathway to determine an equation.
Compare y = 500 * ( 0.94546361966231 )^x with y = 500*e^(kx) aka y = 500*(e^k)^x
We see that
e^k = 0.94546361966231
k = Ln(0.94546361966231)
k = -0.05607986897736
Therefore,
y = 500*e^(kx)
y = 500*e^(-0.05607986897736x)

So we have this candidate list of choices for the equation
y = 500*(0.5)^(x/12.36)
y = 500 * ( 0.94546361966231 )^x
y = 500*e^(-0.05607986897736x)
There are probably other formats to consider.

Personally I prefer the first format because I think it's easiest to work with.
The half-life H = 12.36 is readily visible without having to do any calculations.

After we've determined the equation, we plug in x = 48 to figure out how much potassium-42 is left after 48 hours.
y = 500*(0.5)^(x/12.36)
y = 500*(0.5)^(48/12.36)
y = 33.8782897037247
y = 33.88
or you could say
y = 500 * ( 0.94546361966231 )^x
y = 500 * ( 0.94546361966231 )^48
y = 33.8782897037107
y = 33.88
or
y = 500*e^(-0.05607986897736x)
y = 500*e^(-0.05607986897736*48)
y = 33.8782897036631
y = 33.88
About 33.88 mg of potassium-42 remains after 48 hours.

An informal way to think about it:
1 half-life = 12.36 hrs
1 hr = 1/(12.36) half-lives
48 hr = 48/(12.36) half-lives
48 hr = 3.8835 half-lives approximately
The answer will involve a duration between 3 and 4 half-lives.

We start with 500 mg of potassium-42.
After 12.36 hours (1 half-life), half of it goes away. We have 0.5*500 = 250 mg left.
After 24.72 hours (2 half-lives = 2*12.36 = 24.72 hrs), there will be 0.5*250 = 125 mg left.
After 37.08 hours (3 half-lives = 3*12.36 = 37.08 hrs), there will be 0.5*125 = 62.5 mg left.
After 49.44 hours (4 half-lives = 4*12.36 = 49.44 hrs), there will be 0.5*62.5 = 31.25 mg left.
Therefore, the answer must be between 62.5 mg and 31.25 mg (corresponding to 3 and 4 half-lives).
This method won't nail down the exact value, but it does give an informal way to figure out where the answer should be located.
It can serve as a way to check the answer.

=============================================

Summary:

One equation is y = 500*(0.5)^(x/12.36)
Other equations are possible. See above.

x = number of hours that elapse
y = amount, in mg, of potassium-42 remaining

Approximately 33.88 mg of potassium-42 remains after 48 hours.
Round this value however your teacher instructs.

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