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Week 7 Topic 2 – Exponential Functions
1
Week 7
Topic 2 – Exponential Functions
Reading
There’s a fable from India (you can find various versions online1), with the idea that a
wise person asks the king for a single grain of rice to be placed on the first square of a
chessboard, two grains on the next square, four grains in the square after that, and so on, with
the number of grains of rice doubling with each sequential square.
In every version, the king is surprised by such a simple request. But as you can see in
Figure 1, below, the number of grains of rice has increased 256-fold going just one row down
the chessboard.
Figure 1 – A Chessboard about to be Overwhelmed with Rice
You can grab a calculator and continue to fill in the chessboard, if you like. Once you
have enough numbers to your satisfaction, start reading the next page.
1
One such version is at http://britton.disted.camosun.bc.ca/jbchessgrain.htm
Week 7 Topic 2 – Exponential Functions
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On the 24th square, after the end of the only the third row, there are over 8 million
grains of rice. By this point, the king realized he didn’t have enough rice or wealth to possibly
fulfill his promise.
On the 64th square, there are over 9,000,000,000,000,000,000 grains of rice. That’s over
nine quintillion, or 9⋅1018, grains of rice. Using Wolfram|Alpha’s best estimate for the amount
of rice eaten in the world every year, it’ll take over 500 years to eat all the rice only on the last
square.
In fact, that’s enough rice to stretch to the nearest solar system, Alpha Centauri, and
keep going for another light-year.
This is an example of exponential growth. The numbers get really big, really fast. Our
minds seem better suited to linear thinking. Throwing a ball at a moving target involves
formulas like rate⋅time = distance, a linear relationship.
Let’s make a formula that gives the grains of rice, G, on square N of the chessboard.
Table 1 shows values for most of the first row of the chessboard.
Table 1 – Values of N, G, and the Calculation to get G
Square N
Grains of Rice G
on Square N
Calculation to get G
1
1
1
2
2
1⋅2
3
4
1⋅2⋅2
4
8
1⋅2⋅2⋅2
5
16
1⋅2⋅2⋅2⋅2
6
32
1⋅2⋅2⋅2⋅2⋅2
In the calculation, we multiply the starting amount by 2 an increasing number of times.
Each additional square, we multiply by 2 again. In the last row, N is 6 and we multiply by 2 five
times. On each line, we multiply by 2 N-1 times. For instance, when N = 4, G is 23. This allows us
to write the formula:
G = 1 ⋅ 2 N −1
Check the formula: For N = 6, G = 1 ⋅ 25 , which equals 32 as required. I left the 1 in the
formula because that’s the starting quantity. If the first square had 9.8 grains of rice, the
formula would have 9.8 instead of 1.
Week 7 Topic 2 – Exponential Functions
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We’ve seen equations with exponents before, but this is the first time we have the
variable in the exponent. The exponent here is N − 1 only because of how we set up the
situation. We could rewrite this equation using properties of exponents:
G = 1 ⋅ 2 N −1
G = 1 ⋅ 2 N ⋅ 2 −1
=
1 N
2
2
Go ahead and check this formula to be sure it still works.
This is a specific example, but just as any linear function can be written in slopeintercept form, any exponential function can be written in a general form.
General Form for an Exponential Function
f ( x) = C ⋅ a x
C is the initial amount (when x = 0)
a is called the growth factor
If a>1, the function is increasing
If 0<a<1, the function is decreasing
This is an important formula. You’ll want to memorize it. Any time you’re asked to find
the equation of an exponential function, you’ll go right to this.
C is the initial amount because when I put 0 in for x:
f (0) = C ⋅ a 0
f (0) = C ⋅1
f (0) = C
That’s the Zero Exponent Rule at work: a 0 = 1 for any non-zero value of a .
(In the chessboard example, we didn’t have a square 0, so we ended up with a strange N-1 in
the exponent. However, if we did have a square 0, it would have ½ a grain of rice on it because
when we double ½, we get 1, and there’s 1 grain of rice on square 1. The rewritten formula has
C = ½ to reflect that.)
To find a formula for an exponential function, we need to find values of C and a. If we
know the y-intercept, we know C and can use that to solve for a. But if we don’t know C, we
have a set up and solve a system of equations.
Week 7 Topic 2 – Exponential Functions
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Example 1: Find a formula for the exponential function through (0, 5) and (6, 49).
Solution: We have the y-intercept as (0, 5), so we set C = 5. Our formula now is:
f ( x) = 5 ⋅ a x
Now we put in (6, 49) and solve for a:
49 = 5 ⋅ a 6
49
= a6
5
6
49
=a
5
We don’t need a ± here because only positive values of a make sense in an exponential
function. We could approximate a ≈ 1.4629, but it’s better to leave answers exact. Since 49/5 is
not a repeating decimal, I’ll use a = 6 9.8 in my final formula:
f ( x) = 5 ⋅
(
6
9.8
)
x
For the next few weeks of class (this week, next week on logarithms, and the week after
on applications), you’ll run into many problems where the numbers don’t turn out nice. Usually
you want to give your answers as exact, unless the problem says to round to a specified number
of decimal places. (Dollar values should always be rounded to the nearest penny.)
Example 2: Find a formula for the exponential function through (9, 15) and (12, 25.92).
Solution: We don’t know the vertical intercept, so we use both points to set up a system
of equations:
15 = C ⋅ a 9
25.92 = C ⋅ a12
We can solve the system with substitution. If I solve the first equation for C and put that
into the second equation, I get:
 15 
25.92 =  9  ⋅ a12
a 
1.728 = a 3
1.2 = a
Now put this into either original equation and solve for C:
Week 7 Topic 2 – Exponential Functions
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15 = C ⋅ (1.2 )
9
15
=C
1.29
C ≈ 2.9071, but exact answers are better. It’s difficult to write an exact value of C better
than it’s given above, so I’ll just leave it in that form for my final function:
g ( x) =
15
x
1.2 )
9 (
1.2
Usually one of the two values will turn out nicely, but no guarantees. Figure 2
shows graphs of the functions f and g we found in Examples 1 and 2.
Important Note: The textbook has absolutely no problems where you have to solve for both a
and C, but the Math 111 Supplement (found through Links in D2L) has several of these (with
solutions). Make sure you do those because this type of problem is very common on exams.
Figure 2 – Graphs of the Functions from Example 1 and 2
y
Example 1
Example 2
40
30
20
10
x
-12
-10
-8
-6
-4
-2
2
4
6
8
10
12
The shapes of the graphs are similar. They are both increasing, concave up, and have a
horizontal asymptote at y = 0.
The asymptote appears because if we put a negative value into one of the functions, we
get a negative exponent, which makes the function value very close to zero. However, the yvalues can’t be negative unless the value of C is negative.
You may notice something else about this graph. The function from Example 1 had a
growth factor of about 1.46, which isn’t much bigger than the growth factor of 1.2 from
Example 2, but the graph increases much more rapidly. Figure 3 shows graphs of functions with
different values of the growth factor, and all have C equal to 5.
Week 7 Topic 2 – Exponential Functions
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Figure 2 – Varying Values of the Growth Factor
y
a=0.20
a=1.8
80
60
a=0.75
40
a=1.1
20
x
-15
-10
-5
5
10
15
As you can see, values of a that are very big increase much more rapidly. Values of a
that are close to zero decrease very rapidly. And if you had a = 1, you’d have the constant
function, not increasing or decreasing.
Fun practice! Write formulas for each of these functions. Check by graphing.
Before moving on, let’s make a few more comments about growth factors and
equations. In the chessboard example, the value of G (grains of rice) doubled every time we
went to the next square. That “times two” is a feature of that function the same way the “times
a” is a feature of every exponential function.
For example, Table 2 shows a single value for an exponential function with a = 3.
Table 2 – Fill-in-the-Blanks Exponential Function
-4
-3
-2
-1
0
1
2
3
4
0.9
To fill in other values of this table, we could go through the steps to find the formula
and then put in x = 1, x = 2, x = -4, etc. Or we could use the fact that every time x increases by 1,
the y-value is multiplied by the growth factor, in this case 3.
So when x = -1, y = 0.9⋅3, or 2.7. When x = 0, y = 2.7⋅3 = 8.1. And so on. We multiply by 3
for each step we take to the right.
If we want to take a step to the left, well the opposite of multiplication is division. So
when x = -3, y = 0.9/3 = 0.3. Before moving to the next page, fill in the rest of this table yourself.
Week 7 Topic 2 – Exponential Functions
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Table 3 – The Completed Version of Table 2
-4
-3
-2
-1
0
1
2
3
4
0.1
0.3
0.9
2.7
8.1
24.3
72.9
218.7
656.1
Fun practice! Write a formula for this function. Check by putting in a few x-values.
The point here is that the ratio of successive y-values is equal to the growth factor. If
you want to write the equation of an exponential function through (10, 5500) and (11, 5830),
you can set up the system of equations, substitute, and eventually find a = 1.06. The shortcut is
to realize 11 is one more than 10 (successive x-values), and then divide the y-values:
5830
= 1.06
5500
To put this into function notation:
f (9) f (5) f (n + 1)
=
=
=a
f (8) f (4)
f ( n)
You could make similar statements for ratios of non-successive y-values. For example, in
Table 3, observe that taking two steps to the right multiplies the y-value by 3 twice, or
multiplies by 9. In function notation:
f (10) f (6) f (n + 2)
=
=
= a2
f (8)
f (4)
f (n)
And of course, moving 3 steps to the right multiplies by 27, and 4 steps multiplies by 34,
and so on. So you can really just come up with one rule:
f ( m)
= a m−n
f ( n)
If you ever have a job in financial analysis, or really anything with lots of percentages,
you learn how to work with this fact without even thinking about it.
Percentages and Exponential Growth
This is one of the applications of exponential functions. Technically, we don’t get into
applications for another two weeks, but it makes more sense to me to put the notes here since
we review percentages and talk about exponential growth this week.
If you’re feeling like this is already just too much, you’re welcome to do the homework
based on the notes above and come back to this before you read the notes on the applications.
Week 7 Topic 2 – Exponential Functions
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However, the sections on applications go more in depth, so reading this now might make those
sections easier.
If you recall, the notes on percentages, we saw that you can increase a quantity by r% by
multiplying by (1 + r).
For example, suppose the property tax on your house is $2500 this year, but next year
the taxes are increasing by 3%. The growth rate is 3%, and the growth factor is 1.03. Your tax
next year can be calculated as:
2500 ⋅1.03 = 2575
A $75 increase. For nothing! Outrageous! How are you ever going to get ahead? Well, it
gets worse, because suppose the year after that, your taxes go up another 3%. Rounding to the
nearest dollar, we calculate the next year’s taxes as:
2575 ⋅1.03 = 2652
This is a $77 increase over the previous year. The increases are increasing! Let’s get
ready to really fly off the handle by looking at Table 4, your property tax burden for ten years.
Table 4 – Property Taxes This Year and the Next Ten Years
Years after this
year
Taxes
0
2500
1
2575
2
2652
3
2732
4
2814
5
2898
6
2985
7
3075
8
3167
9
3262
10
3360
($)
Between years 9 and 10, your taxes increase $98. I wonder what will happen if we own
this house for 50 years. To find the taxes in year 50, we could multiply the taxes in year 49 by
1.03, but to find the taxes in year 49 we need to know the taxes in year 48.
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We should probably just write a formula that relates the taxes to the year. Table 5
shows the cumulative calculation of taxes through year 5.
Table 5 – Repeated Application of Growth Factors
Years after this
year
Taxes
0
2500
2500
1
2575
2500⋅1.03
2
2652
2500⋅1.03⋅ 1.03
3
2732
2500⋅1.03⋅ 1.03⋅1.03
4
2814
2500⋅1.03⋅ 1.03⋅1.03⋅1.03
5
2898
2500⋅1.03⋅ 1.03⋅1.03⋅1.03⋅1.03
($)
Total Calculation from Original
This is precisely the situation for which exponents were invented. In year 5 we have:
2500 ⋅1.035
Each additional year, we’ll multiply by 1.03 again, which increases the exponent by 1.
The variable here is years, which we might call t (for time), and since the exponent is changing
with time, that’s where the variable goes. We come up with the formula:
y = 2500 (1.03)
t
The variable y is taxes in year t.2,3 We can evaluate the function for the taxes in year 50:
2500 (1.03)
50
= 10960
Over ten thousand dollars? That may seem insane, but it certainly explains a lot of my
mom’s complaints about house prices. She’s lived through fifty years of small percentage
increases that, over time, have added up to an extraordinary amount.
We can calculate the total percentage increase from year 0 to year 50. We get the 50year growth factor by dividing the two quantities:
10960
≈ 4.384
2500
2
The variable T might make more sense for Taxes, but then our variables would be t and T. No one wants that.
3
It’s unfortunate that the footnote markers look like exponents.
Week 7 Topic 2 – Exponential Functions
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The 4.384 is the growth factor: Taxes have increased more than four-fold. Since the
growth factor is 1 + r, we can say taxes have increased 338.4% over 50 years.
Over another fifty years, taxes would be multiplied by 4.384 once again. We can reduce
this fifty-year growth factor to a one-year:
4.384 = a 50
1.03 = a
The one-year growth factor is 1.03, and the annual percentage increase is 3%, as we
started the example with. Putting exponents on the growth factor takes us to larger time
periods, and root extraction (square roots, cube roots, fiftieth roots) takes us to smaller time
periods. We could even compute an average monthly rate:
1.031/12 ≈ 1.002466
The monthly growth factor is about 1.002466, and the monthly growth rate is 0.2466%.
That’s probably enough about that, for now. Let’s look at another situation. Suppose
you have $100 to invest. You are going to invest in tech stocks, and let’s pretend you are an
exceptional stock trader and get 15% interest each year.
Similarly to the previous example, we can write a formula for the amount of money you
have in the investment account for any given year:
f (t ) = 100 (1.15 )
t
How many years do you think it’ll be before you are a millionaire? Assume you don’t put
any money in and don’t have any fees so your initial $100 just keeps growing.
Unfortunately, it’s going to take you 66 years to get your million. (We’ll solve that
equation once we learn about logarithms.) To speed that up, you could obviously put some
money in each month, but you’d be surprised how quickly the math becomes complicated. The
two things we can change in this formula are the initial amount and the interest rate.
Option 1: Instead of $100, you could invest $10,000 instead.
Option 2: Instead of 15% interest, you could get 35%.
Which of these will make you a millionaire faster? Figure 3 shows graphs of these two
options.
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Figure 3 – Exponential Growth at Work
y
1000000
800000
Millionaire!
600000
Option 1; C = 10,000, a = 1.15
Option 2; C = 100, a = 1.35
400000
200000
x
5
10
15
20
25
30
You can see the initial $10,000 deposit starts higher (well, you can kinda see that on the
graph, but it’s also obviously more money), but the higher interest rate curves upward much
more. Option 2, even though you start with only $100 in the bank, gets you a million bucks
faster.
In fact, an exponential function will give larger y-values more rapidly than any other
function we’ve looked at, in the long run.
The Number e
The textbook introduces the number e in Section 4.3. This is a special number that
appears in many applications in mathematics, just like π.
π ≈ 3.14159265358979
e ≈ 2.71828182845905
The number e continues forever without repeating, which makes it an irrational
number. Other irrational numbers are like 2 . But e is also what’s known in mathematics as a
transcendental number. That’s a number that isn’t a zero of any polynomial function with
rational coefficients. Those are more special.
This number is so special that it deserves proper treatment, but the textbook authors
just casually throw it in without any explanation or reason for it. This course saves e until the
applications where we can indulge a more appropriate motivation for its existence.
When you see e in the textbook before then, just think of it as 2.71828182845905.