Unit 2 – Exponents
Study Guide
7.1 Integer Exponents
Part 1: Zero Exponents
Algebraic Definition:
x 0  1 where x can be any non-zero value
x  0 because 0 raised to any power less
than or equal to zero is an undefined
value.
Example:
40  1
If you look at the chart, when exponents decrease by one in the table
of values, the value is divided by 4, so 40 is equal to 1.
44
256
Power
Value
43
64
4
42
16
4
41
4
4
40
1
4
Exercises: Simplify
1.
12 0 = 1
3.
 4
  = 1
 5
2.
 85
4.
16(0) 0 = undefined
0
= -1
0
Part 2: Negative Exponents
Algebraic Definition:
x n 
1
where x can be any non-zero
xn
value.
x  0 because division by 0 is an
undefined value.
Examples:
x 5 
1
x5
 24 
1
1
1


4
(2) * (2) * (2) * (2) 16
(2)
 2 4  1 *
1
1
1
1
 1 *
 1 *  
4
2*2*2*2
16
16
2
Exercises: Simplify using positive exponents only.
1.
 14
3.
4
 
5
=1
2
=
25
16
2.
a 3b 4 c 2 =
4.

b4
a 3c 2
1 2 4
1
x y =  2 4
2x y
2
7.2 Powers of 10 and Scientific Notation
Part 1: Multiplying by Powers of 10
Example:
61.5 *10 5  61.5 *100,000  6,150,000
12.35 *10 3  12.35 *
1
1
 12.35 *
 12.35  1000  0.01235
3
1000
10
Positive Integer Exponent:
if the exponent is a positive integer, n, move the decimal n
number of spaces to the right
Negative Integer Exponent:
if the exponent is a negative integer, n, move the decimal n
number of spaces to the left
Exercises: Find the value of each expression.
1.
853.4 *105
= 85,340,000
2.
0.163  10 2
= 0.00163
Part 2: Scientific Notation
Example:
A number written in scientific notation has two parts.
Part One has a decimal that is greater than or equal to one and less
than ten.
Part Two is a power of 10.
351,623,463,154 = 3 51,623,463,154 = 3.52 x 1011
decimal
11 spaces after the decimal
= positive power of 10
0.000000001542 = 0.000000001 542 = 1.54 x 10-9
9 spaces before the decimal
= negative power of 10
decimal
Exercises: Write using Scientific Notation.
1.
83,241,256,222,163,001
= 8.32  1016
2.
0.000000210042
= 2.1 107
7.3 Multiplication Properties of Exponents
Property
Algebraic Definition
Example
Product of Powers
a m  a n  a m n
x 4 x 5  x 45  x 9
Power of a Product
abm  a m b m
3x2  32 x 2  9 x 2
Power of a Power
a 
x 
m n
 a mn
 
5 3
 x 53  x15
Examples:
32 * 34 * 32  3242  34  81
4b3  4b4b4b  43 b 3  64b 3
 2x    2x  2x    2 x 
3 2
3
3
2
3 2
 4x 6
Exercises: Simplify fully.
1.
4a bc 4a b c 
2
3
5
4
1
2.
= 16a 7b5c 2
3.
x 2 y 3 * x 3 y 3 * x 5 y 3
= x4 y3
3m 2 (2m 2 ) 2
=
3m 6
4
7.4 Division Properties of Exponents
Property
Algebraic Definition
Examples
Quotient of Powers
 am
 n
a
 k5 
 2   k 52  k 3
k 
 m4 
1
 8   m 48  m 4  4
m
m 

  a m n

m
am
a
Power of a Quotient    m
b
b
3
43
64
4


 
3
125
5
5
2
 x2 
x 22 x 4
 3   32  6
y
y
y 
 x3
 2
y



2

x 6
y4

y 4 x 6
Exercises: Simplify fully.
1.
16 x 3 y 5 z 8
56 x 4 y 5 z 2
=
3.
2.
2z6
7x
 1 
 4
 3x 
2
4.
= 9x8
 4a 3 b 2 


4
 5a b 
256b 4
=
625a 4
4
 2mn 4  m 5 

 3 
4 
 3m n  4n 
m7
=
24n3
2
11.2 Exponential Functions
Part 1: Evaluating Exponential Functions
Algebraic Definition:
f x   ab x , where a  0 , b  1 and b  1
Example:
x
The function, f x   23 , models an insect population after x number of days.
What will the population be after 5 days?
f 5  23  2 * 3 * 3 * 3 * 3 * 3  486 insects after 5 days
5
Exercises:
1.
The function, f x   1500(0.995) x , models a prairie dog population after
x number of years. How many prairie dogs will there be in 8 years?
f 8  1500(0.995)8 = approximately 1441 prairie dogs
2.
The function, f x   80.75 , models the width of a photograph in inches
after it has been reduced by 25% x number of times. What is the width of
the photograph after it has been reduced 3 times?
x
f 3  80.75 = approximately 3.375 inches
3
Part 2: Identifying an Exponential Function
Example:
Determine if the ordered pairs represent an exponential function:
{ ( -2 , ¾ ) , ( -1 , 1 ½ ) , ( 0 , 3 ) , ( 1 , 6 ) , ( 2 , 12 ) }
Place the ordered pairs in a table of values.
**Exponential functions have constant ratios**
the x
values are
increasing
by 1 each
time
x
f x 
-2
3
-1
0
1
2
4
11
2
3
6
12
the y
values are
being
multiplied
by 2 each
time
There is a constant ratio between x values and y values;
therefore, this table represents an exponential function.
Exercises: Determine if the ordered pairs represent an exponential function.
Explain why or why not.
1.
{ ( -1, -9 ) , ( 1 , 9 ) , ( 3 , 27 ) , ( 5 , 45 ) }
No, there is a constant increase in the x-values, but not a constant ratio
between y values.
2.
{ ( -2 , 4 ) , ( -1 , 2 ) , ( 0 , 1 ) , ( 1 , ½ ) }
Yes, there is a constant increase in the x-values and constant ratio of the
y-values. They are being multiplied by ½.
Part 3: Graphing Exponential Functions
Example:
x
Graph:
y  34
Make a table of values and graph the points
x
y
-2
-1
0
1
2
3/16
3/4
3
12
48
Exercises: Graph the following.
1.
y  2x
2.
y  52
x
3.
11.3
1
y  3 
2
x
4.
y  2(0.4) x
Exponential Growth and Decay
Part 1: Exponential Growth
An exponential growth function has the form, y  a1  r  , where:
y = the final/total amount
a = original amount which is greater than 0
r = rate of growth, percentage, written in decimal form
t = time
t
Example:
The original value of a painting is $1400, and the value increases by 9%
each year. Write an exponential growth function to represent this
situation. Find the value of the painting in 25 years.
y = unknown at this time
a = $1400
r = 9% = 0.09
t = 25 years
y  1400(1  0.09) 25
y  1400(1.09) 25
y  $12072.31
Exercises:
1.
In 2000, a sculpture was worth $1200. Its value has been increasing 8%
per year. Write an exponential function to represent the total value of the
sculpture. Find the value of the sculpture in 2010.
y  1200(1  .08)10  approximately $2590.71
Part 2: Compound Interest
nt
 r
A Compound Interest function has the form, A  P1   , where:
 n
A = the Balance after t years
P = Principal / original amount
r = Annual interest rate, written in decimal form
n = number of times interest is compounded per year
t = time in years
Example:
Invest $1000 at an interest rate of 3% compounded quarterly for 5 years
 0.03 
A  10001 

4 

A  1000(1.0075) 20
A  1161.18
A = unknown at this time
P = $1000
r = 3% = 0.03
n = quarterly = 4 times
t = 5 years
Exercises: Find the balance:
1.
$18,000 invested at a rate of 4.5% compounded annually for 6 years
 0.045 
A  180001 

1 

A  $23440.68
2.
6*1
 180001.045
6
$1200 invested at a rate of 3.5% compounded quarterly for 4 years
 0.035 
A  12001 

4 

A  $1379.49
4* 4
 1200(1.00875)16
4*5
3.
$4000 invested at a rate of 3% compounded monthly for 8 years
8*12
 0.03 
A  40001 

12 

A  $5083.47
 4000(1.0025)96
Part 3: Exponential Decay
An exponential decay function has the form, y  a1  r  , where:
y = the final/total amount
a = original amount that is greater than 0
r = rate of decay, percentage, written in decimal form
t = time
t
Example:
The population of a town is decreasing at a rate of 1% per year. In 2000,
there were 1300 people. Write an exponential decay function to model
this situation. Find the population of the town in 2010.
y = unknown at this time
a = 1300
r = 1% = 0.01
t = 10 years
Exercises:
1.
y  1300(1  0.01)10
y  1300(0.99)10
y  1175.7  1176 people
The fish population in a stream is decreasing at a rate of 3% per year. The
original population was 48,000 fish. Write an exponential decay function
to model this situation. Find the total fish population after 7 years.
y  48000(1  0.03)7  48000(.97)7 = approximately 38,783 fish
Part 4: Half Life
A Half Life function has the form, A  P0.5 , where:
t
A = final amount
P = original amount
t = number of half lives in a given time period
Example:
Flourine-20 has a half life of 11 seconds. Find the amount of Flourine-20
left from a 40-gram sample after 44 seconds.
A = unknown at this time
P = 40 grams
t = (44/11) = 4 half lives
A  40(0.5) 4
A  2.5 grams
Exercises:
1.
Find the amount of Flourine-20 left from a 40-gram sample after 2
minutes. Round your answer to the nearest hundredth.
**Remember – Flourine-20 has a half life of 20 seconds **
A = 40(.5)6
2.
= 0.625 grams
Cesium-137 has a half life of 30 years. Find the amount of Cesium-137
left from a 100 milligram sample after 180 years.
A = 100(.5)6 = 1.5625 mg
3.
Bismuth-210 has a half life of 5 days. Find the amount of Bismuth-210
left from a 100 gram sample after 5 weeks.
( HINT: Change weeks to days)
A = 100(.5)7 = 0.78125 g