Download presentation

Presentation is loading. Please wait.

Published byAvis Turner Modified over 2 years ago

1
MODELING REAL SITUATIONS USING EXPONENTIAL FUNCTIONS: PART 1 Patterns #4

2
Example 1 Remember the function A = 1000(1.08) n where $1000 was invested at 8% compounded annually? You were asked when the money doubled. Let’s solve it using logs instead of a graph!

3
Example 1 So A = 1000(1.08) n where $1000 was invested at 8% compounded annually, find out how long before the money doubles. 2000 = 1000(1.08) n

4
Example 1

5
Example 2 Coffee, tea, cola, and chocolate contain caffeine. When you consume caffeine, the percent, P, left in your body can be modeled as a function of the elapsed time, n hours, by the equation: P = 100 (0.87) n Determine how many hours it takes for the original amount of caffeine to drop by 50%.

6
Example 2 So to determine how many hours it takes for the original amount of caffeine to drop by 50% we use the equation: P = 100 (0.87) n. Solve:50 = 100 (0.87) n

7
Example 2

8
Example 3 The population, P million, of Alberta can be modeled by the equation P = 2.28 (1.014) n, where n is the number of years since 1981. Assume this pattern continues. Determine when the population of Alberta might become 4 million.

9
Example 3 The population, P million, of Alberta can be modeled by the equation P = 2.28 (1.014) n, where n is the number of years since 1981. Assume this pattern continues. Determine when the population of Alberta might become 4 million. To get started we need to solve the equation: 4 = 2.28 (1.014) n

10
Example 3

11
Example 4 In 1995, Canada’s population was 29.6 million, and was growing at about 1.24% per year. Estimate the doubling time for Canada’s population growth. We start with an equation: P = 29.6 x (1.0124) n So doubling would mean: 2 x 29.6 = 29.6 x (1.0124) n

12
Example 4

13
Example 5 Consider the equation P = 100 (0.87) n that models residual caffeine. Write this equation as an exponential function with ½ as the base instead of 0.87. To start: write 0.87 as a power of 0.5

14
Example 5

15
In Chemistry… Radioactive isotopes of certain elements decay with a characteristic half-life. You can use an equation of the form:

16
Example 6 In April 1986, there was a major nuclear accident at the Chernobyl plant in Ukraine. The atmosphere was contaminated with quantities of radioactive iodine-131, which has a half-life of 8.1 days. How long did it take for the level of radiation to reduce to 1 % of the level immediately after the accident?

17
Example 6 To begin: Solve for d, the number of days

18
Example 6

19
Example 7 In 1996, Kelowna’s population was approximately 89 000 and was growing at about 3.33% per year. Estimate the doubling time for Kelowna’s population growth.

20
Example 7

21
Example 8 Radioactive phosphorus-32 is used to study liver function. It has a half-life of 14.3 days. If a small amount of phosphorus-32 is injected into a person’s body, how long will it take for the level of radiation to drop to 5% of its original value?

22
Example 8

23
Textwork p. 92/1-12

Similar presentations

OK

Exponential Functions Section 1.3. Exponential Functions What real-world situations can be modeled with exponential functions???

Exponential Functions Section 1.3. Exponential Functions What real-world situations can be modeled with exponential functions???

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on data collection methods quantitative research Ppt on morbid obesity and anaesthesia Ppt on condition based maintenance ppt Ppt on viruses and bacteria worksheet Ppt on operating system overview Free ppt on parts of a flower Ppt on rational numbers Ppt on atm machine download Ppt on recycling of waste in india Ppt on chapter 12 electricity calculator