Laws of Indices or Exponents with Examples

In mathematics indices or index is the value that is raised to the power of any variable or constant like x raised to power 2 i.e. x².

Where x is a variable and 2 is called the power or exponent of that variable.

Let’s see an example:

ax^m

Here

a = coefficient

x =  base

m = power or exponent or index of x

Example:

3x^{5 (power)}

3 = coefficient,

x = base and

5 = power or exponent

When the place of variable (x) deals with numerical values like:

5^m or 6³ 

Here bases are 5 and 6 respectively and powers m and 3 are known as the index (plural form indices).

Laws of Indices

Rule 1

When the base is the same powers will be added.

a^m. aⁿ = a^m+n
Example:

2⁵+2³ = 2⁵⁺³

Rule 2

a^m÷aⁿ = a^m-n

Example 1:

2⁵÷2³

= 2^5-3

= 2²

= 4

Example 2:

3^1/2 ÷3^-1/3

= 3^1/2-(-1/3)

= 3^1/2+1/3

= 3^(3+2)/3

= 3^5/2

Rule 3

(a^m)ⁿ = a^m×n

Example:

(2³)² = 2^3×2

= 2⁶

Rule 4

When the base numerator and denominator interchange with each other then the sign of the exponent will change, + becomes – and – becomes +.

(a÷b)^-m = (b÷c)^m 

or

(a÷b)^m = (b÷c)^-m

Example 1:

(2÷3)^-5 = (3÷2)⁵

Note: a^-m = 1÷a^m

Example 2:

5^-2 = 1÷5²

Rule 5

If the base is different and powers are the same in that situation we can write base as a product.

a^m b^m = (ab)^m

Example:

2³ 5³=(2×5)³

= (10)³

= 1000

Rule 6

If the power of the base is zero, the result is always 1.

a⁰ = 1

√x =x^1/2

Example 1:

√2 = 2^1/2

Example 2:

√16=16^1/2

= (4²)^1/2   

= 4^2×1/2  

Some Practice Questions

Q1. If 8^x+1 = 64, then find x…………?

Solution:

8^x+1 = 64

=> 8^x+1=8² [if the base of both sides is the same then we can compare the powers]

x+1 = 2

=> x=2-1

=> x=1 Ans

Q2. If 5^x = 3125, then find 5^x-3 …….?

Solution:

5^x = 3125

5^x = 5⁵ [ if the base of both sides is the same then we can compare the powers]

x = 5

5^x-3 = 5^5-3 [put x=5]

=> 5^x-3 = 5^5-3

=> 5^x-3 = 5²

=> 5^x-3 = 25

Q3. If 5^x+3=(25)^3x-4, then find x = …….?

Solution:

5^x+3 = (25)^3x-4

=> 5^x+3 = (5²)^3x-4

=> 5^x+3 = 5^2×(3x-4) [(a^m)ⁿ =a ^m×n]

x+3 = 2(3x-4) [if the base of both sides is the same then we can compare the powers]

=> x+3 = 6x-8

=> x-6x = -8-3

=> -5x = -11

=> x = -11/-5

=> x = 11/5

Q4. 27^2x-1 = (243)³, then find x ………..?

Solution:

27^2x-1 = (243)³

=> (3×3×3)^2x-1 = (3×3×3×3×3)³

=> (3³)^2x-1 = (3⁵)³

=> 3^6x-3 = 3¹⁵

[if the base of both sides is the same then we can compare the powers]

6x-3 = 15

=> 6x = 15+3

=>6x = 18

=> x = 18/3

=> x = 6 Ans

Q5. If 6^(2x+1)÷216 =36, find the value of x….?

Solution:

6^(2x+1)÷216 = 36

=> 6^(2x+1)÷216 = 36

=> 6^(2x+1)÷6³ = 6²

=> 6^(2x+1)-3 = 6²  [a^m÷aⁿ = a^m-n]

(2x+1)-3 = 2 [if the base of both sides is the same then we can compare the powers]

=> 2x+1 = 2+3

=> 2x = 5-1

=> x = 4/2

=> x = 2 Ans

Q6. Simplify the following:
(i) 1/(216)^-2/3 + 1/(256)^-3/4  + 1/(243)^-1/5

(ii) (2/3)^-2×(1/5)^-2×(1/6)^-2

Solution:

(i) 1/(216)^-2/3 + 1/(256)^-3/4  + 1/(243)^-1/5

= (216)^2/3 + (256)^3/4 +(243)^1/5        [ 1/a^-m =a^m ]

= (6³)^2/3 + (4⁴)^3/4 +(3⁵)^1/5              [  (a^m)ⁿ=a^m×n  ]

= 6^3×2/3 +4^4×3/4 + 3^5×1/5

= 6² + 4³ +3

=36 + 64+3

= 103

(ii) (2/3)^-2×(1/5)^-2×(1/6)^-2

= (3/2)²×(5/1)²×(6/1)²

= [3/2 × 5×6 ]²

= (45)²

= 2025

When two or more exponents have addition and subtraction signs between them at that time can take common.

Example:

3^x – 3^x+3

=> 3^x – 3^x3³

=>3^x(1-3³)

= 3^x(1-27)

= 3^x(-26)

Q7. If (4⁹⁷-4⁹⁶+4⁹⁵) = k.4⁹⁵ then find the value of k ….?

Solution:

(4⁹⁷-4⁹⁶+4⁹⁵) = k.4⁹⁵ 

=> (4⁹⁵⁺²-4⁹⁵⁺¹+4⁹⁵) = k.4⁹⁵ 

=> (4⁹⁵.4²-4⁹⁵.4+4⁹⁵ ) = k.4⁹⁵ 

=> 4⁹⁵(4²-4+1) = k.4⁹⁵ 

=> 16-4+1 = k

=> 13 = k

Q8. If 2^x-1+2^x+1=320, then find x=…?

Solution:

2^x-1+2^x+1 = 320

[in this situation we will take common in L.H.S. part]

=> 2^x2^-1+2^x2¹ = 320       [ by  a^m. aⁿ=a^m+n]

=> 2^x[2^-1+2¹] = 320

=> 2^x[1⁄2+2¹] = 320

=> 2^x[(1+4)÷2] = 320

=> 2^x[5÷2] = 320

=> 2^x = 320×2⁄5

=> 2^x = 320×2⁄5

=> 2^x = 64×2

=> 2^x = 128

=> 2^x = 2⁷

[If the base is the same, the exponent can be compared]

=> x=7 Ans

Q9. If 2^x+4-2^x+2=3 ,then find x ….?

Solution:

2^x+4-2^x+2 = 3

=> 2^x2⁴-2^x2² = 3

=> 2^x2⁴-2^x2² = 3     [In this situation we will take common in L.H.S. part]

=> 2^x(2⁴ – 2²) = 3

=> 2^x(16-4) = 3

=> 2^x(12) = 3

=> 2^x(12) = 3

=> 2^x = 3⁄12

=> 2^x = 1⁄4

=> 2^x = 1⁄2²

=> 2^x = 2^-2      [If the base is the same, the exponent can be compared]

x = -2 Ans

Q10. If 2^x-2^x-1=16, then find x……?

Solution:

2^x-2^x-1 = 16

=> 2^x-2^x2^-1 = 2×2×2×2

=> 2^x(1-2^-1) = 2⁴

=> 2^x(1-1⁄2) = 2⁴

=> 2^x(2-1)/2 = 2⁴

=> 2^x1⁄2 = 2⁴

=> 2^x = 2⁴2⁄1

=> 2^x = 2⁵              [If the base is the same, the exponent can be compared]

=> x = 5 Ans

I hope you understood about laws of indices or exponents. If still, you have any doubts, you can ask in the comment section below.