Textbook, Chapter 5, Sections 5-1 through 5-7 (pages 169–207)
Section 5-1: Growth and Decay: Integral Exponents pp. 169–175
Let's review work with exponents
This last problem showed that 30 = 1. It is a fact that b0 = 1. In problem 8,
the simplification has a
negative exponent. That is because there were more 3s in the denominator than in
the numerator . For
that reason, the end result has the 27 in the denominator. If you compare the
results of problems 7
and 8, you see they are the inverse of each other. So b-3 is the inverse of b3.
The inverse of b3is
When a problem is completely simplified, there are no negative
exponents or powers of powers .
Example 1: Simplify
You cannot add unless exponents and bases are identical. So simplify 3-2
and 32 first:
Simplify the numerator by squaring −4 and multiplying the exponents:
Simplify the denominator:
Divide and subtract the exponents:
Exponents are used to find values that increase or decrease exponentially.
This occurs when an item
increases or decreases by a set percentage every year. If the cost of a car
increases yearly by 8%,
next year it would cost 100% of this year's cost plus an additional 8%. Then the
year after that, it
would increase by another 8%. So every year the new price could be found by
multiplying by 1.08,
the decimal for 108%. A function can be used to find the new price in t years:
A0 is the beginning amount, r is the rate of increase as a decimal, and t is
the number of years the
price has been increasing. This function can also be used to find the price
years in the past, if a
negative t is used.
Example 5: A car now costs $4,000. Car prices are
increasing at 10% a year. (a) What will the price
be in 6 years? (b) What was the price 5 years ago?
Example 6: Let's try problem 38 on page 174 of your
Example 7: Let's try problem 48 on page 175 of your text:
If 2x+1 was 2x, the problem could be factored. If I
divide 2x+1 by 2 , it will be
Replace 2x with y and factor:
Replace y with 2x:
Complete odd-numbered problems 1–49 in the Written Exercises on pages 173–175
of your text.
Then check your answers in the back of the text.
Section 5-2: Growth and Decay: Rational Exponents pp. 175–180
As briefly mentioned in the previous section, roots are fractional exponents.
Other than that, all
previous rules hold .
Example 1: Writein
Example 2: Simplify
Example 3: Let's try problem 26 on page 178 of your text:
Example 4: Solve
They have the same bases so the exponents must be the same:
Example 5: Solve
Raise both sides by a power of −1/2:
Example 6: Let's try problem 52 on page 179 of your text.
There is a greatest common factor of
Example 7: Let's try problem 38 on page 179 of your text:
Divide by 140:
Raise both sides to the power of 1/4:
The annual percent of increase has been 6.7%.
Complete odd-numbered problems 1–57 in the Written Exercises sections on
pages 178–180 of your
text. Then check your answers in the back of the text.
Section 5-3: Exponential Functions pp. 180–186
The general form for an exponential function is f (x) = abx. With this form,
function values can be
given and the function can be found. Given f (0) = 4 and f (3) = 32, the
function can be found using
We have used A (t) = A0(1 + r)t for exponential growth and decay
when the rate of change is known.
If the rate is not known,
can be used. k is the time needed to
multiply A0by b.
Example 1: Let's try problem 8a on page 183 of your text:
The value of b is 1/2 because the half-life is the time at which half of the
substance will remain. The
value of k is 1,600 because it takes 1,600 years for 1/2 of it to remain. The
value of t is 3,200 because
we want to see how much remains after 3,200 years:
After 3,200 years 1/4 of a kilogram will remain.
Example 2: Let's try problem 16 on page 184 of your text:
There will be .000977 times the original amount of medicine in the blood
stream after 30 days.
Example 3: Let's try problem 12 on page 184:
The population in 2,000 will be 40 million.
(b) The rule of 72 states an estimate for doubling time can be found by
dividing 72 by the rate.
So 72/3 = 24. It would take approximately 24 years to double.
Example 4: One last problem: problem 28 on page 186 of your text.
(a) The population doubled from 100 million to 200 million in 60 years. So
the doubling time is
60 years. Using
(b) After trying 30 years, 36 years, 33 years, and 35 years as t values in
the equation above, I
have concluded that t = 35 years gives a population of 299.66 million. This is
very close to
Complete odd-numbered problems 1–29 in the Written Exercises on pages 183–186
of your text.
Then check your answers in the back of the text.
Section 5-4: The Number e and the Function ex pp. 186–190
The letter e has a special meaning in math, statistics, and physics. It is
the value of the natural
exponential function at 1. The letter e was chosen for this value in honor of
its discoverer, Leonhard
Euler. Its value can be approximated using
As x gets increasingly larger, the value of y
approaches 2.7182818. This process of placing larger and larger values in for x
can be shown using
calculus terminology. The limit
terminology means you can keep placing values in for
n until you reach infinity.
The value of e is used in this section to find annual yield on a savings
account that has interest compounded daily . The function
can be used for any item that grows continually. Your
calculator should have an ex key to help you work with this function.
Example 1: Suppose you invest $1.00 at 8% interest. How much would you
have at the end of the
year if interest compounds (a) quarterly, (b) monthly, (c) continuously?
Quarterly means four times a year. So the 8% interest will be paid in 2%
would have 102% of your money every 3 months:
Monthly means 12 times a year. So the 8% interest will be paid in .667%
would have 100.667% of your money every 12 months:
For continuous compounding, use
The value of r is the interest rate as a
decimal, and t is the time in years:
Example 2: Let's try problem 12 on page 189 of your text.
The problem states t represents number of days from now. So to find the number
present now, t = 0.
There are 3,000 ladybugs now.
7 days in a week:
In a week, there will be 3,217 ladybugs.
Example 3: Let's try parts (a) and (c) in problem 20 on page 190 of
When h = 1, point A is (0, f (0))and point B is (1, f (1)).
When h = .01, point A is (0, f (0))and point B is (.01, f (.01)).