Question

Suppose the data below are the result of a study of a medication used to fight a certain bacteria. The x column represents the number of hours after the medication was introduced to the bacteria and the y column represents the number of bacteria present. Use the data in the table below to answer each question.

Answer 1

1. What does the scatterplot of the data show?
a. A strong positive linear relationship
b. A strong negative linear relationship
c. A curve that represents exponential growth
d. A curve that represents exponential decay

2. Complete an exponential transformation on the y-values. What is the new value of y when x = 5?
a. 1.301
b. 0.699
c. 1.4472
d. 1.1761

3. Find the linear transformation model.
a. Log y^ (y-hat) = -12.1667x + 84.8333
b. Log y^ (y-hat) = 84.8333x – 12.167
c. Log y^ (y-hat) = -0.1375x + 1.9978
d. Log y^ (y-hat) = 1.9978x – 0.1375

4. Predict the number of bacteria present for 12 hours.
a. 0.35
b. 2.23
c. -61
d. 0

Answer 2

I know that the first one is d, but I don't know the others or how to find them. If anyone could explain how to get the answers, that would be great because I am confused.

Answer 3

It certainly looks like d: http://www.wolframalpha.com/input/?i=transpose+ {{0%2C1%2C2%2C3%2C4%2C5%2C6%2C7}%2C{100%2C73%2C53%2C38%2C28%2C20%2C15%2C11}}

Answer 4

You might need to copy and paste into your browser.

Answer 5

If this is an exponential model, you'll need to find two parameters, an initial value \(P_0\) and the decay rate \(a\) so then \(y=P_0e^{-at}\). \((\star)\)

To do this, simply use any two <x,y> values. For example, at x=1, y=73 and at x=3, y=38. So,
$$
\large
73=P_0e^{-a(1)} \implies a=-\ln\left (\cfrac{73}{P_0}\right )\\
$$
Then, using that decay rate, \(a\), and the second point (<3,38>):
$$
\large
38=P_0e^{-a(1)} \implies P_0=\cfrac{38}{e^{-a(3)}}
$$

Once you have this formula, you can find the linear transformation by taking logs on both sides of \((\star )\)

Make sense?

Answer 6

The second equation with y=38 should have been 3:
$$
38=P_0e^{-a(3)}
$$

Answer 7

Here are the solutions for \(P_0\) and \(a\): http://www.wolframalpha.com/input/?i=73%3Dx*exp%28-y%29%2C38%3Dx*exp%28-3*y%29

Answer 8

It makes a bit more sense, knowing that I need to use the formula and plug stuff in, but I'm still confused. I don't know anything about log or what the different things in the formula mean. I'm behind; I've been sick so I need to catch up. I don't understand though. Would you mind explaining what the formula means and what log is? I'm sorry about asking this much of you but I need help.

Answer 9

FYI, based on the last link I sent, \(P_0=102\) and \( a=0.33\). So,
$$
y=102e^{-0.33t}
$$
Taking logs of both sides, we get:
$$
\log(y)=\log(102e^{-0.33t})\\
\log(y)=\log(102)-0.33tlog(e)\\
\log(y)=2-0.14t
$$
Which is c) in problem 3.

So this may not make much sense if you don't understand logs.

What do you know about logs?

Answer 10

For problem 2, just plug in x=5 into our transformation:
$$
y=2-.14(5)=1.3
$$
Which is a)

Answer 11

Should have been:
$$
\log(y)=2-.14(5)=1.3
$$

Answer 12

The last question is the easiest. Just plug in 12 for \(t\) in \((\star)\).

Want to go over logs?

Answer 13

Thank you so much for all of the help. I am understanding so much better now. I really appreciate it. I'm new to this site, how do you give medals? Because you definitely deserve one.

Answer 14

Yes, I would like to go over logs if you don't mind.

Answer 15

just click "Best Response" to give medals

Logs were invented by those guys that didn't have calculators way back a long time ago, so to make it easy to multiply large numbers, they created log tables.

For example y=log(2500*1000)=log(2500)+log(1000)

Then to get 2500*1000, all you had to do was take your answer and plug into \(10^y\).

So the rule is \(\log(a\times b)=\log a+ \log b\).

Make sense so far?

Answer 16

so if log(y) = 2 - 0.14t
then for number 4 it would be: log(y) = 2 - 0.14(12) which would be 0.32, yes? and the answer closest to that would be 0.35 so the answer would be a.

Answer 17

Check this. \(\log (100)=\log(10\times10)=\log 10+\log 10\). Take a calculator, type "100" then "log" write down your answer. Next, take your calculator and type "10" then "log" and then add to "10" then "log". This sum will equal to the answer you just wrote down.

Answer 18

Yes, that makes sense so far.

Answer 19

So take the example we just used, \(\log 100\). This can be rewritten as \(\log 10^2\).
Right?

Answer 20

because 100 = \(10^2\)

Answer 21

Okay, that makes sense.

Answer 22

Notice that \(\log(100)=\log 10^2=\log(10)+\log10\)
and that \(\log(1000)=\log 10^3=\log(10)+\log(10)+\log10\)
So the exponent determined the number of times we need to add \(\log(10)\). This is no accident.

The rule is \(\log10^N=N\log(10)\). If we are using a base of 10, then log(10) = 1. The "1" means \(10^?=1\). The ? is equal to 1, that's what log(10) means.

So log(100)=2, which means 10^2=100.

If we are using a different base, say base \(e\), then we write log as \(ln\), which is called the natural log. So, \(\ln e^2=2\) and \(\ln e^3=3\), etc.

See that?

Answer 23

*\(10^?=10\), not \(10^?=1\) above in second paragraph.

Answer 24

These rules above are the basic ones for handling logs.

Why transform?

You transform when you have an exponential curve and you want to make it linear.

So, if you have exponential data, taking it's natural log, \(\ln\), makes it linear. When you have "linear" data, it's a straight line when graphed. People like straight lines because then you can predict what the value will be at some other point-in-time much easier, just extrapolate.

Once you have extrapolated and know what the value will be, you "undo" the transformation by exponentiating your answer.

That's what we did above, we found the value of the transformed value in question 2 after finding the "linear" function that resulted from taking log of our \(\star\) equation in question 3.

Answer 25

Oh, That makes so much sense! I get it.

Answer 26

Answers are:
1. D
2. A
3. C
4. B