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May 18th

May 18th

# Exponents and Logarithms

PURPOSE

This lesson concentrates on simplifying expressions containing exponents and working with
exponential functions. This also leads to the inverse of the exponential function, the logarithmic
function.

OBJECTIVES

After completing this lesson, you should be able to:
•simplify expressions containing negative exponents ;
•simplify expressions containing fractional exponents ;
•define and use exponential functions;
•define and apply the natural exponential function;
•define and apply logarithms;
•apply laws of logarithms; and
solve exponential equations and change logarithms from one base to another.

Textbook, Chapter 5, Sections 5-1 through 5-7 (pages 169–207)

COMMENTARY
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
Thus 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:

Example 2: Simplify

When raising a power by a power, multiply the exponents :

Example 3: Simplify

Distribute the monomial :

Add exponents of like bases :

Invert terms with negative exponents:

Example 4: Simplify

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 text.

Change all bases to powers of 2:

Change all bases to powers of 3:

The square root of a number is also a power of 1/2:

Change all bases to powers of 5:

The cube root of a number is also a power of 1/3:

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 2x:

Replace 2x with y and factor:

Replace y with 2x:

Study Exercises

Complete odd-numbered problems 1–49 in the Written Exercises on pages 173–175 of your 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 exponential form:

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%.

Study Exercises

Complete odd-numbered problems 1–57 in the Written Exercises sections on pages 178–180 of your

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
substitution:

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
300 million.

Study Exercises

Complete odd-numbered problems 1–29 in the Written Exercises on pages 183–186 of your 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?

a)
Quarterly means four times a year. So the 8% interest will be paid in 2% increments. You
would have 102% of your money every 3 months:

(b)
Monthly means 12 times a year. So the 8% interest will be paid in .667% increments. You
would have 100.667% of your money every 12 months:

(c)
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.

(a)
The problem states t represents number of days from now. So to find the number of ladybugs
present now, t = 0.

(b)
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 your text.

(a)
When h = 1, point A is (0, f (0))and point B is (1, f (1)).

(c)
When h = .01, point A is (0, f (0))and point B is (.01, f (.01)).

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