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 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 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 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 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
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
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 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
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