Multiplication Properties of Exponents

Monomial A number, variable, or the product of a number and

one or more variables.

Constant Any monomial that is a real number.

Expression Monomial ? Reason

-5 yes Constant, real number

p + q no The plus sign

x Yes variable

𝑐

𝑑 no variable denominator

𝑎𝑏𝑐8

5 yes =

1

5𝑎𝑏𝑐2

• Exponents or Powers

Be careful with negative bases:

−(𝟐)𝟒 = −𝟏 ∙ 𝟐 ∙ 𝟐 ∙ 𝟐 ∙ 𝟐 = −𝟏𝟔 (−𝟐)𝟒 = −𝟐 ∙ −𝟐 ∙ −𝟐 ∙ −𝟐 = 𝟏𝟔

EXPONENT RULES GRAPHIC ORGANIZER

Name Rule Examples

Adding & Subtracting Monomials

COMBINE LIKE TERMS!!

(Do Not Change common variables and exponents!)

1. 9𝑥2𝑦 − 10𝑥2𝑦 = −1𝑥2𝑦

2. Subtract 6w from 8w 8w-6w=2w

Product Rule

𝑥𝑎 ∙ 𝑥𝑏 = 𝑥𝑎+𝑏

Keep the base, add exponents

1. ℎ2 ∙ ℎ6 = ℎ8

2. (−2𝑎2𝑏) ∙ (7𝑎3𝑏) =

−14𝑎5𝑏2

Power Rule

(𝑥𝑎)𝑏 = 𝑥𝑎∙𝑏

Keep the base, multiply exponents

1. (𝑥2)3 = 𝑥6

2. (−2𝑚5)2 ∙ 𝑚3 =

4𝑚13

Quotient Rule

𝑥𝑎

𝑥𝑏= 𝑥𝑎−𝑏

Keep the base subtract exponents

1. 27𝑥5

42𝑥=

27𝑥4

42

2. (𝑦2)2

𝑦4 =𝑦4

𝑦4 = 1

Negative

Exponent Rule

𝑥−𝑎 =1

𝑥𝑎

1

𝑥−𝑎= 𝑥𝑎

1. −5𝑥−2 =−5

𝑥2

2. 4𝑘2

8𝑘5 =1

2𝑘3

Zero Exponent

Rule

𝑥0 = 1

1. 7𝑥0 = 7(1) = 7

2. (𝑤4)2

𝑤8 =𝑤8

𝑤8 = 1

You Try:

Let’s practice the Product Rule, Power Rule and Zero Exponent Rule!

Simplify x3 ∙ x2 ∙ x6 = x3+2+6

= x11

Simplify 4x4y7 ∙ 2xy5 ∙ x6

= (4)(2)(x4+1+6)(y7+5) = 8x11y12

Simplify (a7b8)5 = a(7)(5)b(8)(5)

= a35b40

Simplify (3x2y6z3)4

= 34x(2)(4)y(6)(4)z(3)(4)

= 81x8y24z12

Simplify (2m3n2)2 ∙ (m2n2)3 = 22m(3)(2)n(2)(2) ∙ m(2)(3)n(2)(3)

= 4m6n4 ∙ m6n6 = 4m6+6n4+6 = 4m12n10

Simplify 8m5n4 ∙ (2mn3)2

= 8m5n4 ∙ 22m(1)(2)n(3)(2)

= 8m5n4 ∙ 4m2n6 = 32m5+2n4+6 = 32m7n10

Simplify 3x0y3z6 ∙ 7y0z2 = 3(1)y3z6 ∙ 7(1)z2

= (3)(7)y3z6+2 = 21y3z8

Simplify (9a7b3c12)0

= 1

Division Properties of Exponents

Quotient of Powers

When dividing monomials

with the same base, We subtract the

Exponents Divide or simplify coefficients!

𝑥5

𝑥3= 𝑥5−3 = 𝑥2

------------------------- 8𝑦9

2𝑦3= 4𝑦6

Power of a Quotient

Find the power of the numerator and the

power of the denominator, then

subtract.

(𝑥

𝑦)3 =

𝑥3

𝑦3

--------------------------

(3𝑥2

4𝑥)2 =

9𝑥4

16𝑥2=

9

16𝑥2

For more complicated division problems line up like variables before applying

the rules:

6𝑥5𝑦3𝑧2

2𝑥𝑦𝑧

= 3𝑥4𝑦2𝑧

3𝑎4𝑏2

6𝑎3

=1𝑎𝑏2

2

Negative Exponents: We need to “fix” any nonzero number raised to a negative

exponent.

When solving problems with negative exponents, circle the variable with the

negative exponent and move just that variable and negative exponent to the

other side of the fraction bar and the exponent becomes positive.

NOTE: NEGATIVE COEFFICIENTS DON’T GET MOVED UNLESS THEY ALSO

HAVE A NEGATIVE EXPONENT!!

−2𝑎−1𝑏2

𝑐−3= −

2𝑏2𝑐3

𝑎

You Try!

Simplify 7𝑎6𝑏4𝑐2𝑑5

21𝑎4𝑑2

= 1𝑎2𝑏4𝑐2𝑑3

3

Simplify 24𝑚12𝑛6

8𝑚8𝑛

= 3𝑚4𝑛5

Simplify ( 5𝑥3

15𝑥2)2

= 25𝑥6

225𝑥4

= 𝑥2

9

Simplify 𝑎7𝑏3

𝑎4𝑏2𝑐−3

= 𝑎3𝑏𝑐3

Simplify 15𝑥5𝑦−2

𝑧−4

= 15𝑥5𝑧4

𝑦2

Simplify 𝑎7𝑏2𝑐4

𝑎3𝑏5𝑐4

= 𝑎4𝑏−3𝑐0

= 𝑎4

𝑏3

Rational Exponents

Radical Sign When there is no number in the elbow, we assume ‘2’

√25 = 5*5

√83

is read cube root of 8 and is equal to 2 * 2* 2

Exponential expression → Rational Expression

62

3 → (√63

)2 The denominator of the fraction is the

elbow of the radical and the numerator is the exponent.

Let’s go from exponential to rational and rational to exponential:

5𝑥1

2 = 5√𝑥

(5𝑥)1

2=√5𝑥

Solve Exponential Equations: In an exponential equation, the variable is an

exponent. To solve, we must make sure both sides of the equal sign have the

same base – then we can set the exponents equal to each other.

If 5x = 57 then x = 7. If 2x = 8 then 2x = 23 and x = 3

You Try:

6𝑥 = 216

6𝑥 = 63

𝑥 = 3

82x = 16

(23)2x = 24

6x = 4 Power to a power

x = 4

6=

2

3

25𝑥−1 = 5

(52)𝑥−1 = 51 Power to a power

52(𝑥−1) = 51

2(𝑥−1) = 1

2𝑥−2 = 1

2𝑥 = 3

𝑥 =

27x-5 = 9

27x-5 = 32

x – 5 = 2

x = 7

1

9

Exponential Functions

Exponential Functions Non-linear

𝑦 = 𝑎𝑏𝑥 𝑤ℎ𝑒𝑟𝑒 𝑎 ≠ 0 𝑎𝑛𝑑 𝑏 ≠ 1

• The exponent is the variable

𝑦 = 𝑎𝑏𝑥 + 𝑐 (c value indicates range)

Graph 𝑦 = 3𝑥

x y

-2 (𝑦 = 3−2)

-1 1

3

0 1

1 3

2 9

y-int=1 domain = all real numbers range = {y Į y>0} c = 0

Exponential Growth Exponential Decay

𝒚 = 𝒂𝒃𝒙 𝒚 = 𝒂𝒃𝒙 Growth Decay

a > 0, b > 0 a > 0, 0 < b < 1

(b is a decimal or fraction)

Domain: All real numbers Domain: All real numbers

Range: Positive Real Numbers Range: Positive real #s.

Identifying Exponential Behavior:

Method 1 Method 2

Domain (x) regular intervals. (+5)

Range (y) common factor (1/2)

YES! Exponential Yes! Exponential Graph!

x 0 5 10 15 20 25 y 64 32 16 8 4 2

Graph 𝑦 = (1

4)𝑥 . Find the y-intercept and domain and range.

x y -2 16 -1 4 0 1 1 1

4

2 1

16

y-int = 1

domain= all real #

range = {y Į y>0}

Graph 𝑦 = 2(3)𝑥 + 1. Find the y-intercept and domain and range.

x y -2 1

2

9

-1 12

3

0 3 1 7 2 19

y-int = 3

domain= all real #

range = {y Į y>1}

Growth and Decay

𝑦 =𝐶(1+𝑟)𝑡

𝑦 =𝐶(1−𝑟)𝑡

𝐴 =𝑃(1+𝑟

𝑛)nt

Y = final amount

C = Initial Amount r

= rate of change

(as decimal) t = time

Y = final amount

C = Initial Amount r

= rate of change

(as decimal) t = time

A = current amount

P = Principal - Initial Amount

r = rate of interest (as decimal) t =

Number of years

n= Number of times compounded

Exponential

Growth

Exponential

Decay

Compound

Interest

Examples:

1. The prize for a radio station contest begins with $100 gift card. Once a day a

name is announced. The person has 15 minutes to call or the prize increases by

2.5% for the next day.

a. Write an equation to represent the amount of the gift card in dollars after t

days with no winners.

b. How much will the gift card be worth if no one wins after 10 days?

a. 𝑦 = 𝑐(1 + 𝑟)𝑡 (since the amount grows, we use growth

formula)

𝑦 = 100(1 + .025)𝑡 (plug in what you know. C=100; r → 2.5% =

.025)

𝑦 = 100(1.025)𝑡 equation that represents amount of the gift

card in dollars after t days with no winners.

b. 𝑦 = 100(1.025)𝑡 (Substitute 10 for t and solve.)

𝑦 ≈ 128.01

In 10 days, the gift card will be worth $128.01

2. Maria’s parents invested $14,000 at 6% per year compounded monthly. How much

money will be in the account after 10 years?

𝐴 = 𝑃(1 + 𝑟

𝑛)𝑛𝑡

What do we know: P = 14,000

r = 6% or .06

n = 12 (monthly) t = 10

𝐴 = 14000(1.005)120

𝐴

𝐴 ≈ 25471.55

There will be about $25,471.55 in the account after 10 years.

Geometric Sequences as Exponential Functions

Geometric Sequence A sequence in which each term after the nonzero

first term is found by multiplying the previous

term by a constant called the common ratio (r). r

≠ 0 𝑜𝑟 1.

Remember

When we looked at arithmetic sequence, we looked for a common difference

between the terms.

Well

For geometric sequence we look for a common ratio. (HINT: we can do this by

dividing the last term by the one before it and repeat)

a) 1, 4, 16, 64, 256 … Common Ratio = 4, yes this is geometric

sequence

b) 1, 3, 5, 7, 9 ….

This is NOT a geometric sequence because

it has a common difference not

a common ratio! Therefore it is

arithmetic sequence

c) 4, 9, 12, 18…

This is neither a geometric sequence nor an

arithmetic sequence.

To find the next few terms of a geometric sequence:

Step 1: Find the common ratio

Step 2: Multiply the last term by the common ratio to get the next term

Find the next three terms in the geometric sequence:

20, -28, 39.2, _____, _____, _____

Common ratio = = −1.4

Next three terms: (39.2)(-1.4) = -54.88

(-54.88)(-1.4) = 76.832

(76.832)(-1.4) = -107.5648

nth term: just like we were able to find the nth term of the arithmetic

sequence using a formula, we can find the nth term of a

geometric sequence:

𝒂𝒏 = 𝒂𝟏 ∙ 𝒓𝒏−𝟏 where 𝑎1= first term in sequence

a. Find the ninth term for the sequence -6, 12, -24, 48

𝑎1 = −6 𝑛 = 9 𝑟 = −2

𝒂𝟗 = −𝟔 ∙ (−𝟐)𝟗−𝟏

= −𝟔(−𝟐)𝟖

= −𝟏𝟓𝟑𝟔

b. Write an equation for the nth term:

𝒂𝒏 = −𝟔(−𝟐)𝒏−𝟏

Recursive Formulas

Recursive Formula Allows you to find the nth term of a sequence

by performing operations to one or more of

the preceding terms.

USING RECURSIVE FORMULAS:

Substitute the first term into the given equation to find the 2nd term.

Substitute the 2nd term into the given equation for 𝑎𝑛−1 to find the 3rd term.

Repeat until you’ve found all the terms asked for:

Find the first five terms of the sequence in which:

𝑎1 = 7 𝑎𝑛 = 3𝑎𝑛−1 − 12 if n≥ 2

First term given equation

𝑎1 = 7 given

𝑎2 = 3(7) − 12 = 9

𝑎3 = 3(9) − 12 = 15

𝑎4 = 3(15) − 12 = 33

𝑎5 = 3(33) − 12 = 87

Sequence: 7, 9, 15, 33, 87…

WRITING A RECURSIVE FORMULA:

Step 1: Step 2: Step 3:

Arithmetic

17, 13, 9, 5, …

Common difference = -4

𝑎𝑛 = 𝑎𝑛−1 + 𝑑

𝑎𝑛 = 𝑎𝑛−1 + (−4)

𝑎𝑛 = 𝑎𝑛−1 − 4

𝑎1 = 17 𝑎𝑛 = 𝑎𝑛−1 − 4 𝑛 ≥ 2

Geometric

6, 24, 96, 384 …

Common Ratio = 4

𝑎𝑛 = 𝑟 ∙ 𝑎𝑛−1

𝑎𝑛 = 4 ∙ 𝑎𝑛−1

𝑎1 = 6 𝑎𝑛 = 4𝑎𝑛−1 𝑛 ≥ 2

Step 1: Determine if sequence is arithmetic or geometric – find

common difference or common ratio.

Step 2: Arithmetic → 𝑎𝑛 = 𝑎𝑛−1 + 𝑑

Geometric → 𝑎𝑛 = 𝑟 ∙ 𝑎𝑛−1

Step 3: State the first term and domain for n