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. x^{2}.

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 6^{3 }

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}. a^{n }= a^{m+n}**

**Example:**

2^{5}+2^{3 }= 2^{5+3}

### Rule 2

**a ^{m}÷a^{n} = a^{m-n}**

**Example 1:**

2^{5}÷2^{3}

= 2^{5-3}

= 2^{2}

= 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:**

(2^{3})^{2 }= 2^{3×2}

= 2^{6}

### 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÷5^{2}

### 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:**

2^{3 }5^{3}=(2×5)^{3}

= (10)^{3}

= 1000

### Rule 6

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

**a ^{0}^{ }= 1**

**√x =x ^{1/2}**

**Example 1:**

√2 = 2^{1/2}

**Example 2:**

√16=16^{1/2}

= (4^{2})^{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}=8^{2 }[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 }= 5^{5} [ 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} = 5^{2}

=> 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 }= (5^{2})^{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}

=> (3^{3})^{2x-1 }= (3^{5})^{3}

=> 3^{6x-3 }= 3^{15}

[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)}÷6^{3 }= 6^{2}

=> 6^{(2x+1)-3 }= 6^{2 }[a^{m}÷a^{n} = 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} ]

= (6^{3})^{2/3} + (4^{4})^{3/4} +(3^{5})^{1/5} [ (a^{m})^{n}=a^{m×n} ]

= 6^{3×2/3} +4^{4×3/4} + 3^{5×1/5}

= 6^{2} + 4^{3} +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^{x}3^{3}

=>3^{x}(1-3^{3})

= 3^{x}(1-27)

= 3^{x}(-26)

**Q7. If (4 ^{97}-4^{96}+4^{95}) = k.4^{95 }then find the value of k ….?**

**Solution:**

(4^{97}-4^{96}+4^{95}) = k.4^{95 }

=> (4^{95+2}-4^{95+1}+4^{95}) = k.4^{95 }

=> (4^{95}.4^{2}-4^{95}.4+4^{95 }) = k.4^{95 }

=> 4^{95}(4^{2}-4+1) = k.4^{95 }

=> 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^{x}2^{-1}+2^{x}2^{1 }= 320 [ by a^{m}. a^{n}=a^{m+n}]

=> 2^{x}[2^{-1}+2^{1}] = 320

=> 2^{x}[1⁄2+2^{1}] = 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 }= 2^{7}

[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^{x}2^{4}-2^{x}2^{2 }= 3

=> 2^{x}2^{4}-2^{x}2^{2 }= 3 [In this situation we will take common in L.H.S. part]

=> 2^{x}(2^{4 }– 2^{2}) = 3

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

=> 2^{x}(12) = 3

=> 2^{x}(12) = 3

=> 2^{x }= 3⁄12

=> 2^{x }= 1⁄4

=> 2^{x }= 1⁄2^{2}

=> 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^{x}2^{-1 }= 2×2×2×2

=> 2^{x}(1-2^{-1}) = 2^{4}

=> 2^{x}(1-1⁄2) = 2^{4}

=> 2^{x}(2-1)/2 = 2^{4}

=> 2^{x}1⁄2 = 2^{4}

=> 2^{x }= 2^{4}2⁄1

=> 2^{x }= 2^{5 }[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.