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Exponential growth and decay were introduced in Algebra 1.
In Algebra 2, we will review and expand these concepts.
The following two exponential formulas can be used to illustrate the concepts of exponential growth and decay in applied situations. If a quantity grows (or decays) by a fixed percentage at regular time intervals, the pattern can be depicted by these functions.
Exponential Growth:
y = a(1 + r)x
The growth factor is greater than 1. |
a > 0, (1 + r) > 1
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Exponential Decay:
y = a(1 - r)x
The decay factor is less than 1. |
a > 0, 0 < (1 - r) < 1
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Remember that the original exponential formula was y = abx.
You will notice that in these growth and decay functions,
the b value (growth factor) has been replaced either by (1 + r) or by (1 - r).
The growth "rate" (r) is determined as b = 1 + r.
The decay "rate" (r) is determined as b = 1 - r
"r" is the percentage of growth or decay
a = initial value (the amount before measuring growth or decay)
r = growth or decay rate (most often represented as a percentage and expressed as a decimal)
x = number of time intervals that have passed
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Remember that exponential growth describes a quantity that is increasing by a fixed percentage at regular (fixed) time intervals.
Example 1: The population of HomeTown is 2016 was estimated to be 35,000 people with an annual rate of increase of 2.4%.
a) What is the growth factor for HomeTown?
The "growth factor" refers to b as represented by (1 + r),
Since this problem deals with "increase", b = (1 + r) where r = 2.4% = 0.024
The growth factor 1 + r = 1 + 0.024 = 1.024. (Remember that growth factor is greater than 1.)
b) Write an equation to model future growth.
y = abx = a(1 + r)x = a(1.024)x = 35000(1.024)x
c) Use the equation to estimate the population in 2020 to the nearest hundred people.
There are 4 annual "time intervals" from 2016 to 2020, so x = 4.
y = 35000(1.024)4 ≈ 38,482.91 ≈ 38,500 |
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Remember that exponential decay describes a quantity that is decreasing by a fixed percentage at regular (fixed) time intervals.
Example 2: The number of dolphin sitings in Baseline Cove, when first recorded, were 250 per year, but have been decreasing by 3% per year since that first siting.
a) What is the decay factor for Baseline Cove sitings?
The "decay factor" refers to b as represented by (1 - r),
Since this problem deals with "decrease", b = (1 - r) where 3% = 0.03
The decay factor 1 - r = 1 - 0.03 = 0.97. (Remember that decay factor is between 0 and 1.)
b) Write an equation to model future decay.
y = abx = a(1 - r)x = a(0.97)x = 250(0.97)x
c) Use the equation to estimate the population of dolphins sited 5 years after the first siting. Round to nearest whole dolphin.
There are 5 annual "time intervals" for sitings in a 5 year period (x = 5).
y = 250(0.97)5 ≈ 214.6835064 ≈ 214 dolphins |
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| Continuous Growth or Decay: |
Most naturally occurring phenomena grow continuously. For example, bacteria will continue to grow over a 24 hour period, producing new bacteria which will also grow. The bacteria do not wait until the end of the 24 hours, and then all reproduce at once.
The exponential e is used when modeling continuous growth that occurs naturally such as populations, bacteria, radioactive decay, etc.
You can think of e like a universal constant representing how fast you could possibly grow using a continuous process. And, the beauty of e is that not only is it used to represent continuous growth, but it can also represent growth measured periodically across time (such as the growth in Example 1).
In Algebra 2, the exponential e will be used in situations of continuous growth or decay.
The following formula is used to illustrate continuous growth and decay.
Continuous Exponential Growth or Decay
A = ending value (amount after growth or decay)
A0 = initial value (amount before measuring growth or decay)
e = exponential e = 2.71828183...
k = continuous growth rate (also called constant of proportionality)
(k > 0, the amount is increasing (growing); k < 0, the amount is decreasing (decaying))
t = time that has passed |
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If we compare this new formula to our previous exponential decay formula (or growth formula),
we can see how ek is related to the rate of decay, r, (or growth). |
Example 3: A chemical sample with an initial mass of 30 grams is decaying at a continuous rate of 7.2% per year. How many grams, to the nearest tenth, will remain after 4 years?
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The rate of continuous decay will be a negative value
(k < 0). So, k = -7.2%. Fortunately, we have sufficient information to fill in the entire right side of the equation. This will make the solution a simple computation of the right side. |
Example 4: The current enrollment in the nursing program at a local community college is 500 students. The enrollment is increasing continuously at a rate of 1.5% each year. Which logarithm is equal to the number of years it will take for the enrollment to increase to 600?
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Substitute into the equation:
A0 = 500
A = 600
k = 0.015
To solve for t, we need to isolate the e0.015t.
Take ln of both sides, using the inverse relationship on the right.
Choice 4.
FYI: This is approximately 12.2years.
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Example 5: The number of bacteria in a Petri dish is continuously increasing. This situation is modeled by the exponential function N = 80e3t, with N = the number of bacteria present in the Petri dish after t hours. Find, to the nearest hundredth, the number of hours it will take to reach 2000 bacteria.
N = N0 ekt
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You may see the general formula written with different letters.
Since the value of k is positive 3, this is continous growth.
Substitute 2000 for N (the desired end result) and solve for t.
Use the natural log (ln) to solve for t, since it is the inverse operation when working with e.
Round answer to nearest hundredth.
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To see more about continuous growth and decay,
go to Doubling and Half-Life and Financial Interest.
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