Exponents

Learning Outcomes:

After reading the post readers will be able to:

1) Recognize base, exponent and value.

2) Know about the product law of exponents.

• aˣ × aⁿ = aˣ⁺ⁿ
• aⁿ × bⁿ = (ab)ⁿ

3) Quotient law.

• aˣ ÷ aⁿ = aˣ⁻ⁿ
• aⁿ ÷ bⁿ = (a/b)ⁿ

4) power law. (aˣ)ⁿ = aˣⁿ

5) For zero(0) exponent: a⁰ = 1

6) For exponent as negative Integer. a⁻ⁿ = (1/aⁿ)

7) Concept of power of integer that is  (-a)ⁿ when "n" is even or odd integer.

Exponents or Indices:

Base, Exponent and Value:

We know that the repeated multiplication of n number can be written in a short by using exponent.

 For example

5×5×5 can be written as 5³. We read it as 5 to the power of 3 where 5 is the base and 3 is the exponent or index.

Note: The exponent of a number indicates that, how many times a number is multiplied with itself.

Similarly as above, 12×12 can be written as 12² . We read it as 12 to the power of 2 where 12 is the base and 2 is the exponent.

So from above examples we can conclude that if a number "s" is multiplied with itself n-1 times then the product will be sⁿ ,i.e.

Sⁿ = s×s×s×s×.......×s  (n-1 times multiplication of s with itself)

We read it as "s to the power of n" or "nth power of s" where "s" is the base and "n" is the exponent.

Example: Express in exponential form. (-5) × (-5) × (-5)

Solution: (-5) × (-5) × (-5) = (-5)³

Example: Express in exponential form. 3×3×3×3×3×3×3

Solution: 3×3×3×3×3×3×3 = (3)⁷

Example: Express in exponential form. (1/7)×(1/7)×(1/7)×(1/7)

Solution: (1/7) × (1/7) × (1/7) × (1/7) = (1/7)⁴

Example: (-3/8) × (3/8) × (-3/8) 

Solution: (-3/8) × (-3/8) × (-3/8) = (-3/8)³

Example: Identify the exponent and base 34⁵

Solution: 34⁵

Base = 34

Exponent = 5

Example: Identify exponent and base of (-3/8)⁷

Solution:

(-3/8)⁷

Base = (-3/8)

Exponent = 7

Example: Identify exponent and base (-x/z)ᵅ

solution (-x/z)ᵅ

Base = (-x/z)

exponent = a

Example: write in simplest form (-2)⁵

Solution: (-2)⁵

= (-2)×(-2)×(-2)×(-2)×(-2)

= -32 Ans.

Example: (2/3)³

Solution: (2/3)³

= (2/3)×(2/3)×(2/3)

=8/27 Ans.

Example: write in simplest form (-1/4)⁴

Solution: (-1/4)⁴

= (-1/4)×(-1/4)×(-1/4)×(-1/4)

= 1/ 256 

Note: [ (-) × (-) = (+) ] ,  [ (-) × (+) = (-) ] and [ (+) × (+) = (+) ]

Laws of exponents OR Indices:

Exponents are used to solve many problems so it is important that we understand the laws for working with exponents. Now we discuss these laws one by one.

Product Law:

1) when bases are same but exponents(power) are different:

See the example to understand

2³ × 2² = (2×2×2) × (2×2)

= (2×2×2×2×2)

= 2⁵

From this example we can notice that same result can be obtained by adding the exponents of two numbers.

2³ × 2² = 2³⁺² = 2⁵

Similarly,

(-3/7)² × (-3/7)⁵ = (-3/7)×(-3/7)×(-3/7)×(-3/7)×(-3/7)×(-3/7)×(-3/7)

= (-3/7)⁷

Again using the short method to find result

(-3/7)² ×(-3/7)⁵ = (-3/7)²⁺⁵ = (-3/7)⁷

Form above we can deduce the following law:

While multiplying two(2) numbers with the same base we add their exponents but the base remains unchanged i.e. for any number "b" with exponents x and n, this law is written as

(b)ˣ × (b)ⁿ = (b)ˣ⁺ⁿ

2) when bases are different but exponents are same:

As we know

2³ × 5³ = (2×2×2) × (5×5×5)

= (2×5) × (2×5) × (2×5)

= (2×5)³

Similarly,

(-1/7)³ × (3/7)³ 

= [ (-1/7)×(-1/7)×(-1/7) ] × [ (3/7)×(3/7)×(3/7)]

= (-1/7 × 3/7)(-1/7 × 3/7)( -1/7 × 3/7)

= ( -1/7 × 3/7)³

We can deduce the law from above examples

While multiplying two(2) rational numbers having the same exponent the product of two(2) bases I written with the given exponent. Suppose two rational numbers are "x" and "y" with exponent "n" then  xⁿ × yⁿ = (xy)ⁿ

Example: Simplify (-1/4)² × (2/3)²

Solution:

(-1/4)² × (2/3)² = [ (-1/4) × (2/3)]²

= [ -2/12 ]² = [ -1/6 ]² Ans.

Example: (-3/2)³ × (-3/2)⁴

Solution : (-3/2)³ × (-3/2)⁴

= (-3/2)³⁺⁴ = (-3/2)⁷ Ans.

Quotient Law:

1) when bases are same but exponents are different

See the following examples to understand

2⁷ ÷ 2³ = 2⁷/2³

= (2×2×2×2×2×2×2)/ (2×2×2)

= 2×2×2×2 = 2⁴

Find the same quotient by another method

2⁷/2³ = 2⁷⁻³ = 2⁴

Similarly,

(-3/7)⁵ ÷ (-3/7)² 

= [ (-3/7)×(-3/7)×(-3/7)×(-3/7)×(-3/7)] ÷ [(-3/7)×(-3/7)]

(-3/7) ×(-3/7)×(-3/7) = (-3/7)³

Short method to find Quotient is

(-3/7)⁵ ÷ (-3/7)² = (-3/7)⁵⁻² = (-3/7)³

We can suggest another law from above examples

The division of two rational numbers with same base can be performed by subtracting their exponents. Let suppose "b" is the base of any two rational numbers with exponents "x" and "n" such that" b" is not equal to zero and "x" is greater than "n" then,

bˣ ÷ bⁿ = bˣ⁻ⁿ

2) when bases are different but exponents(power) are same

As we know

(2/3)⁴ = (2/3)×(2/3)×(2/3)×(2/3)

= (2×2×2×2)/ (3×3×3×3)

= 2⁴/3⁴ = 2⁴ ÷ 3⁴

Similarly,

(x/y)⁵ = (x/y)×(x/y)×(x/y)×(x/y)×(x/y)

= ( x⁵ / y⁵) or x⁵ ÷ y⁵

This law can be written as

If two rational numbers "a" and "b" where "b" is not equal to zero and "n" is there exponent, then

aⁿ ÷ bⁿ = (a/b)ⁿ

Example: 9⁸ ÷ 3⁸ simplify it.

Solution: 9⁸ ÷ 3⁸  = (9/3)⁸ = 3⁸

Example: (-3/11)⁷ ÷ (-3/11)⁴

Solution: (-3/11)⁷ ÷ (-3/11)⁴

= (-3/11)⁷⁻⁴ = (-3/11)³

Example: (14)¹¹ ÷ (63)¹¹

Solution: (14)¹¹ ÷ (63)¹¹ = (14/63)¹¹ = (2/9)¹¹

Power law:

As we studied that bᵅ × bⁿ = bᵅ⁺ⁿ. Let we use this law to simplify an expression (7⁴)².

(7⁴)² = 7⁴ × 7⁴ = 7⁴⁺⁴ = 7⁸ is the same as 7⁴ˣ²

We solve another expression by using same law

[ (-1/2)⁷ ] = (-1/2)⁷ × (-1/2)⁷

= (-1/2)⁷⁺⁷ = (-1/2)¹⁴ is the same as (-1/2)⁷ˣ²

From above examples we deduced that the base remains same with a new exponent equal to the product of the two (2) exponents, i.e.

(bᵅ)ⁿ =bᵅˣⁿ= bᵅⁿ

Zero exponent:

By quotient Law , anything divided by itself is 1 as below.

4²/4² = (4×4) / (4×4) = 1

This can also b written as

4²/4² = 4²⁻² = 4⁰ = 1

Similarly,

(-5)⁴ / (-5)⁴ = 1

Or it can be written as also

(-5)⁴ / (-5)⁴ = (-5)⁴⁻⁴ = 1

So any non zero rational number with zero exponent is equal to 1. Suppose "z" be any non zero rational number with zero exponent then

z⁰ = 1

Negative Exponents:

Look at the given pattern below.

10² = 10 × 10

10¹ = 10

10⁰ = 1

10⁻¹ = 1/10

10⁻² = 1/10 × 1/10

......... so on

and 

10⁻ⁿ = 1/[10×10×10×........×10(n-1 times multiplication)] =1/10ⁿ

In general it can be written as  

a⁻ⁿ = 1/aⁿ

Example: Express in a single exponent. (3⁴)⁵

Solution: 

(3⁴)⁵ = 3⁴ˣ⁵ = 3²⁰

Example: [(-2/3)³]²

Solution:  [(-2/3)³]² = (-2/3)³ˣ² = (-2/3)⁶

Example: [(1/7)⁵]⁶ 

Solution:

[(1/7)⁵]⁶ = (1/7)⁵ˣ⁶ = (1/7)³⁰

Example: change following negative exponent into positive exponent.

1) (-2/5)⁻⁴

2) (a/-b)⁻⁶

Remember that,

When we multiply a negative number by itself it gives us positive result because minus(-) time minus is plus(+).

 For example

(-3)×(-3) = (+9) and (-7)×(-7) = (+49)

Now see an other pattern given below

(-2)² = (-2)×(-2) = (+4) 

(-2)³ = (-2)×(-2)×(-2) = (-8)

(-2)⁴ = (-2)×(-2)×(-2)×(-2) = (+16)

(-2)⁵ = (-2)×(-2)×(-2)×(-2)×(-2) = (-32) ....so on.

From above examples it is clear that a negative number with an odd exponent gives a negative result. So it can b explain as,

Let "x" be any positive rational number and "n" be any non zero Integer, then according to this law:

1) If "n" is an even integer, then (-x)ⁿ is positive.

2) If "n" is an odd integer, then (-x)ⁿ is negative.

Example: (2/5)⁻³ × (2/5)³ + (3/5)⁵ × (3/5)⁻⁵

Solution:

 (2/5)⁻³ × (2/5)³ + (3/5)⁵ × (3/5)⁻⁵

= (2/5)⁻³⁺³ + (3/5)⁵⁺⁽⁻⁵⁾

= (2/5)⁰ + (3/5)⁰ = 1+1 = 2 Ans

Example: (-2/7)⁵ × (-2/7) × [(-2/7)²]

Solution:

(-2/7)⁵ × (-2/7)⁻² × [(-2/7)²]⁻¹

= (-2/7)⁵⁺⁽⁻²⁾ × (-2/7)²ˣ⁽⁻¹⁾

= (-2/7)³ × (-2/7)⁻²

= (-2/7)³⁺⁽⁻²⁾

= (-2/7)³⁻² = (-2/7)¹ = -2/7 Ans.

Have a nice day.