Nate Hansen 03.15.13. Blog Post.

Nate Hansen 03.15.13. Blog Post. 

Hi Everyone, I am blogging for just Friday, as Thursday was filled with Pi day festivities. On Friday, we went over the previous night's homework, and worked with a worksheet Lisa handed out, titled "Exponential Modeling". 

 

I will first go over some of the trickier problems from the homework that was discussed during Friday's class (as a reminder, the homework for that day can be found on the worksheet that is called "Exercises 5.2", and the problems were: #s 60,  51, 53, 54):

 

60. The figure is the graph of an exponential growth function f(x) = Pa^x.

            (a) In this case, what is P? [Hint: What is f(0)?]

            (b) Find the rule of the function f by finding a. [Hint: What is f(2)?]

The first step to solving this problem is finding the value of P. 

As the problem hints, plug in 0 for x, and the resultant value will be the initial y-value, or P (one should also note that the starting y-value is always the y-intercept).

After examining the graph of the function, we find that when x = 0, y = 4, so P = 4.

In order to complete part (b) of the problem, we will need to plug in all known values to find the value of a. After plugging in the known values, the function should look like this:

36 = 4(a)²

By solving this equation, we find that a = 3. By extension, the rule of this function is: f(x) = 4(3)^x.

54. An eccentric billionaire offers you a job for the month of September. She says that she will pay you 2¢ on the first day, 4¢ on the second day, 8¢ on the third day, and so on, doubling your pay on each successive day.

 

(a)   Let P(x) denote your salary in dollars on day x. Find the rule of the function P.

(b)   Would you be better off financially if instead you were paid $10,000 per day? [Hint: Consider P(30).]

 

For those visual learners, you can take a look at the chart below to see the growth pattern of this function:

Number of Days	Money in Dollars
0	.01
1	.02
2	.04
3	.08
4	.16
5	.32
6	.64
7	1.28
8	2.56
9	5.12
10	10.24
11	20.48
12	40.96
13	81.92
14	163.84
15	327.68
16	655.36
17	1310.72
18	2621.44
19	5242.88
20	10485.76
21	20971.52
22	41943.04
23	83886.08
24	167772.16
25	335544.32
26	671088.64
27	1342177.28
28	2684354.56
29	5368709.12
30	10737418.24

 

As shown by the above chart, the amount paid on the first day of work is .02 dollars and theoretically the value of f(0) is half of f(1) because the payment grows by a factor of 2 daily, so the value of f(0) is .01.

From this knowledge, and the fact that the function doubles every day, we find the rule of the function:

P(x) = 0.01(2)^x

To answer part (b) of this problem, we can plug in 30 for x to represent the last day of the month of September, and get a y-value of 10737418.24. Because this number and the cumulative payments for the previous days far exceeds the “$10,000 a day” figure, we know that it would make more financial sense to be paid .02 dollars on the first day of the job, and to double this amount every day afterwards for the month of September. 

The final problem I will do from Friday’s class is problem #4 on the worksheet entitled “Exponential Modeling”.

Tell whether the function represents exponential growth or decay and tell the rate of growth or decay for each.

4. j(x) = 5(4)^x÷9

Because of our knowledge of exponent rules and basic arithmetic, this problem is fairly simple. We first rewrite the problem like this:

j(x) = 5(4)^(1÷9)(x÷1)

Then, because of the “power-to-power” rule, we simplify this function so that:

j(x) = 5(4^(1÷9))^(x÷1)

In this way, we find that this function grows by 1.16652904…(simply the simplification of 4 to the one ninth), or about 17% after x time. 

 If anyone is looking for more resources to learn about Exponential Equations, check out this Khan Academy video:

Thats it for my blog post. Hope it helps. 

Nate