MODELING REAL SITUATIONS USING EXPONENTIAL FUNCTIONS: PART 1

Patterns #4

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!

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

Example 1

log2000 log1000

log1.08n

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%.

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

Example 2

log50 log100

log0.87n

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.

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

Example 3

log4 log2.28

log1.014n

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

Example 4

log2

log1.0124n

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

Example 5

log0.87

log0.5n

P 100(0.5)x

5

In Chemistry…

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

P I(0.5)n

y

where :

P is the percent remaining

I is the initial amount

n is the time elapsed

y is the length of a half - life

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?

Example 6

To begin:

Solve for d, the number of days

1100(0.5)d

8.1

Example 6

1100(0.5)d

8.1

1

1000.5

d

8.1 or 0.010.5d

8.1

log0.01log0.5d

8.1 or log0.01d

8.1log0.5

8.1log0.01

log0.5d

d 53.81523514 or about 54 days

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.

Example 7

log2

log1.0333n

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?

Example 8

5 100(0.5)d

14.3

5100

0.5d

14.3 or 0.05 0.5d

14.3

log0.05 log0.5d

14.3 or log0.05 d

14.3log0.5

14.3log0.05log0.5

d

d 61.80357176 or about 62 days

Textwork

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