Question

This table shows the mass in grams m of the radioactive substance iodine-131 remaining in a container t days after the beginning of an experiment.

Answer 1

Write a function that models the data.

Answer 2

This problem is a bit more challenging than the 3 before it in that you are not told what kind of model to use.

Answer 3

You could try a linear model (which would probably be a poor fit to the data), a quadratic model or an exponential model.

Answer 4

Could I use a quadratic? Or would I use something else?

Answer 5

I could try doing a quadratic.

Answer 6

Let me summarize those 3:

y=ax + b
y=ax^2 + bx + c (our old friend)
y=ae^(bx) exponential model

Answer 7

I had some help on this question from a friend, he told me to use y=ae^(bx), but I didn't know how to use it.

Answer 8

Remember, we don't always get Perfect Fits when we develop models.
Its very possible you could create an acceptable expo model or an acceptable quad model.

Answer 9

I believe I've seen this question before and that the other person and I came up with an exponential model.

Basically, you're doing the same thing, Ricky: finding suitable values for the constants a, or a and b, or a and b and c.

Answer 10

In this case, with 2 constants undetermined, we'd need only 2 points from the table to determine those 2.

Answer 11

Please go ahead and pick 2 pts from the table. Typel them here as (x,y).

Answer 12

(1,917.40), (2,841.62)

Answer 13

Then I substitute for x and y right?

Answer 14

Fine. Now stick those values into y=ae^(bx).

for exampel, from the first point, 841.62 = a e^(b*2), where x = 2 and y =841.62.

To answer your qeustion, yes.

Answer 15

Why don't you use your point (2, 841.62) but use ( 0 , 1000) for the 2nd?

You'll see why in a moment.

Answer 16

Using these 2 points, please come up with 2 equations in a and b.

Answer 17

i can do that!

Answer 18

Much like you did earlier when trying to find a, b and c for a quadratic model.

Answer 19

sounds wonderful, Ricky: "I can do that!" more power to you.

Answer 20

so this is what I have so far:


1000 = ae^(b(0)) -> ae = 1000
841.62 = ae^(b(2)) -> ae^(2b) = 841.62

Answer 21

The first one is actually ae^0 = 1000. Note that e^0 is 1, whereas e =2.718 approx.
Big difference.

Answer 22

So,. solve 1000 ae^0 for a.

Answer 23

The value of a is 1000.
Then yo ur expo model becomes y=1000e^(bx)

Substitute the data from the other point: x=2 and y=841.62.
This is all the info you need to solve for b.

Answer 24

And if that is right (which it is), then i could do this to solve for a:

1a = 1000 -> 1000


Then I can do this to solve for b:

(1000)e^(2b(2)) = 841.62


correct?

Answer 25

Exactly. Think: what's the easiest way to isolate b?

Answer 26

think of properties of log and expo functions.

Answer 27

Well first you would have to multiply 2b and 2 together to get 4b. Then after that, would you divide?

Answer 28

Are you u sing properties of logs and expo functions here?

Answer 29

Oh, no. I wasn't

Answer 30

841.62=1000e^(2b)

your best bet is to take the natural log (ln) of boths ides.

ln 841.62 = ln 1000 + ln e^(2b)

Hope this looks familiar for you.
ln 841.62-ln 1000 = 2b.

Before we do any more be certain that what I've demo'd here is clear for y ou. If it isn't we'll need to talk.

Answer 31

Yes, it is clear!

Answer 32

ln 841.62-ln 1000 = 6.735 - 8.294 = 2 b

Answer 33

I'd forgotten that you don't have a calculator. How are you computing your logs?

Answer 34

b = 417.306

Answer 35

I don't have graphing calculator like yours, but I do have one that does log functions.

Answer 36

6.735 - 8.294 = 2 b => all you have to do is to divide both sides of this equation by 2, and you'll have your b. Note that b is qute different from 417.3.

Answer 37

try it, please.

Answer 38

I come up with the exponential model

y=1000 e ^ (-0.0862 x).

To check this, simply pick another point from the table of x and y values, substitute them into this model and determine whether the resulting equation is true or false.

Answer 39

I got:

-0.7795 = b

Answer 40

You are doing the right thing. Why your b and my b are different, I don't know. I was just beginning to check my work.

Answer 41

While I do that plese think of any last min. questions you'd liek to ask. I have to excuse myself very soon as I am dri8ving towards Los Angeles this morning for an 11:30 a.m. apopiontment.

Answer 42

okay! Let me see.

Answer 43

I tried the same work again, same approach, and still get b = -0.0862.

Answer 44

hmm, okay. Well, that was the last one that I really needed help on! Thank you so much! I think I can take it from here. If i have a question, I will just post it on here for whenever.

Answer 45

Try this

If 841.62=1000 e^(bx),

841.62
------ = e^(bx)
1000

Taking the log of both sides (after re-writing 841.62/1000 as 0.8416),

ln 0.8416 = 2b

b=ln 0.8416/2 = ?

Answer 46

-0.865598

Answer 47

I should have enclosed that log in parentheses for added clarity.

b=(ln 0.8416)/2 = ?

Answer 48

-0.0862252

Answer 49

OK, I think we've spent enuf time on that one.
Very happy to work with you! Probably see you not today but tomorrow or later this week.

Answer 50

There you go. Correct result.

Answer 51

Thank you! Have a great day!

Answer 52

Then your expo model is y = 1000 e^(-0.0862x).

Answer 53

Bye!