Topic 1.3c_ MATHX402-032 Math for Management
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Feb 20, 2024
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10/9/23, 6:52 AM
Topic 1.3c: MATHX402-032 Math for Management
https://onlinelearning.berkeley.edu/courses/2072192/pages/topic-1-dot-3c?module_item_id=96528771
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Topic 1.3c
Return to Module 1
(https://onlinelearning.berkeley.edu/courses/2072192/pages/module-01)
Topic 1.3c Using Exponential functions to Model
Growth and Decay
These 5 generations of Bunnies show a pattern of Exponential growth. How do you know? Click to
reveal.
The pattern 2, 4, 8, 16, 32 is a pattern of sequential growth of exponents on the base 2: , , ,
, .
Continuous growth and the use of e
Population growth and growth of money are common application problems where exponential and
logarithmic functions are used. Sometimes, this growth is discrete
(periodic, like, say, when interest
compounds once per year). Sometimes, the growth is continuous
(at a constant rate, like, say,
growth of bacteria, or bunnies). Continuous growth is modeled using , an irrational number, like ,
that has a value of about 2.718. Why? Briefly, it turns out that the limit, as gets very large, of the
expression:
...is , or 2.718. Go ahead, try it on your calculator, putting in increasingly large values for .
Remember our model of exponential growth of money, or compound interest, from Topic 1.3a is:
As we shorten the compounding periods (to, say, quarterly, monthly, daily, hourly…) and increase the
number of periods in the model (approaching continuous growth), the model is given as:
10/9/23, 6:52 AM
Topic 1.3c: MATHX402-032 Math for Management
https://onlinelearning.berkeley.edu/courses/2072192/pages/topic-1-dot-3c?module_item_id=96528771
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Logically, you can see in this form, how the rate, , would get smaller as the compounding periods
shorten, and the number of periods, , would get very large, now resembling the basic formula
for the value of . At its limit (i.e., when approaches infinity), the value of the continuously growing
population, or continuously compounding principal, becomes:
I hope this has given you a theoretical basis for why is used in models of continuous growth, in
finance and otherwise. Don’t worry if you aren’t 100% clear on the calculations we’ve shown here.
You'll see applied examples here, in your textbook, in later modules (Module 8) and in your later
studies, that will make these equations, and the intuition behind them, more accessible to you.
Example 1.3c.1
How long will it take for $5,000 to grow to $10,000 at 8%, per year, compounded annually?
Because interest is paid annually, we use our interest compounding model for discrete growth:
Filling in the parts that we know:
Now, because we need to solve for a variable in the exponent, we know we need to use logs. We can
either:
(a) take the log of both sides:
And apply the Log of a Power property:
(b) or convert to log form:
And apply the change of base formula:
Check your answer by putting 9 back in the equation for :
10/9/23, 6:52 AM
Topic 1.3c: MATHX402-032 Math for Management
https://onlinelearning.berkeley.edu/courses/2072192/pages/topic-1-dot-3c?module_item_id=96528771
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And check that it works!
Example 1.3c.2
In 2007, Canada’s population was 33,390,141. How many years will it take to double the population if
growth is 0.9% per year?
First, let’s recognize that this is a model of continuous
growth. Using our model , we fill
in what we know. Because we are seeking the years it will take the population to double
, our equation
looks like:
Now, take the natural log of both sides:
years
Q: Why did we take the natural
log of both sides? Click to reveal.
A: Because we are modeling continuous
growth, and our equation uses base .
Example 1.3c.3
Let’s now take the case where the base of the exponential function is between zero
and one (
), the case of exponential decay.
Let’s say that the value of a car is given by .
Q: What is the current value of the car?
A: $6,000
Q: What is the rate of depreciation?
A: We can see that our base is not , so we can assume that this valuation model is a discrete one.
Using the model for discrete growth/decay , we can deduce that if
, then , indicating a depreciation rate of 18%.
Q: What will be the value of the car after 5 years?
A: Substituting 5 in for , we get:
Summary
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10/9/23, 6:52 AM
Topic 1.3c: MATHX402-032 Math for Management
https://onlinelearning.berkeley.edu/courses/2072192/pages/topic-1-dot-3c?module_item_id=96528771
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In this section, we clarified the distinction between discrete and continuous growth, and how this
dictates our approach to applied problems involving exponential growth and decay. Specifically, we
introduced the use of the number e, an irrational number of approximately 2.718, that is used for
modeling continuous growth. Then, we used what we have learned so far in this section, in
application problems of exponential growth and decay.
Return to Module 1
(https://onlinelearning.berkeley.edu/courses/2072192/pages/module-01)