Exponential Growth and Decay

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The_Original_81

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I'm having trouble with 2 problems, each on exponential growth and decay.

1. A student is trying to determine the half-life of radioactive iodine-131. He measures the amount, A, of iodine-131 in a sample solution every 8 hours. His data are shown in the table.

Time (h) 0, 8, 16, 24, 32, 40, 48
Amount (g) 5.79, 5.57, 5.47, 5.25, 5.17, 5.01, 4.85

(a)Find an appropriate exponential model of the data points. Amount at time t, A(t)=????
(b)Find the half-life of idoine-131 according to the exponential model in part (a). Half-life, t=????

2. The doubling period of a baterial population is 15 minutes. At time t=110 minutes, the baterial population was 80000. (a) What was the initial population at time t=0? (b)Find the size of the baterial population after 5 hours.
 
I figured out the 2nd question, I just need the first. I would show what work I've done but I don't even know how to approach it.
 
1. A student is trying to determine the half-life of radioactive iodine-131. He measures the amount, A, of iodine-131 in a sample solution every 8 hours. His data are shown in the table.

Time (h) 0, 8, 16, 24, 32, 40, 48
Amount (g) 5.79, 5.57, 5.47, 5.25, 5.17, 5.01, 4.85

(a)Find an appropriate exponential model of the data points. Amount at time t, A(t)=????
(b)Find the half-life of idoine-131 according to the exponential model in part (a). Half-life, t=????
Click to expand...

If one puts the data into a calculator and does an exponential regression, the result is

A(t) = (5.76626)(.996436)^t

This is an exponential function of the form y = ab^x. However, this is NOT what you are looking for.

You want an equation in which the “b” term is ½ and the exponent is t/k, where k is the half-life: y = a(.5)^(t/k)

“t” is the total elapsed time, and “k” is the half-life period, so the ratio of t/k represents the number of half-lives that have elapsed. Make sense so far?

Here is a quick method that will get you a reasonably good approximation of k.

Let’s discuss “a”. Notice what happens when t=0; we get (.5)^(0/k) = 1, so we find that y = a. This means that “a” is the y-intercept. From the data set, we see that the y-intercept is (0,5.79), meaning a = 5.79.

So now our equation looks like this:

A(t) = (5.79)(.5)^(t/k)

That leaves only k to be found. Let’s choose the point furthest away from the (0,5.79) point to improve our accuracy. Use (48,4.85). Plug this into our equation and solve for k:

4.85 = (5.79)(.5)^(48/k)

Can you take it from here?
 
My HW is online, and I entered the first equation A(t) = (5.76626)(.996436)^t and it was correct. To find the half life, you need k?
 
wjm11 said:
To find the half life, you need k?
Click to expand...

k IS the half-life.
Click to expand...

Wow I sound stupid lol. But to be honest, my teacher never did a problem like this in class. The half life was always given in our problems. And I've already taken the test on this subject and the decay question had to do with carbon dating. I got a 95 on that test so it's not like I don't know what I'm doing. I even looked in the textbook for a problem similar to this and it wasn't there. That's why it's so frustrating. :(
 
To find k using the first equation, set it equal to the other equation form:

A(t) = (5.76626)(.996436)^t
A(t) = (5.79)(.5)^(t/k)

Let's simplify it a little by pretending the 5.766 is equal to the 5.79 and canceling them out. Then set the equations equal to each other:

(.996436)^t = (.5)^(t/k)

Now solve for k. (Hint: make the t terms go away.)
 
wjm11 said:
To find k using the first equation, set it equal to the other equation form:

A(t) = (5.76626)(.996436)^t
A(t) = (5.79)(.5)^(t/k)

Let's simplify it a little by pretending the 5.766 is equal to the 5.79 and canceling them out. Then set the equations equal to each other:

(.996436)^t = (.5)^(t/k)

Now solve for k. (Hint: make the t terms go away.)
Click to expand...

How would you do that? Is it an exponential equation?
 
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