EXPONENTS AND SQUARE ROOTS

Exponent :

The exponent of a number says how many the number has to be multiplied by itself.

In 9² the '2' says that 9 has to be used twice twice in multiplication, so 9² = 9 × 9 = 81.

In words : 9² can be called '9 to the power 2' or '9' to the second power, or simply '9 squared' Exponents are also called Powers or Indices.

Square Root :

Square root of a number is a value that can be multiplied by itself to give the original number. 

The symbol of the square root is

√

Square root of 9 is 3. 

Because when 3 is multiplied by itself, we get 9.

Exponent Rules

Rule 1 : 

x^m ⋅ xⁿ  =  x^m+n

Rule 2 : 

x^m ÷ xⁿ  =  x^m-n

Rule 3 : 

(x^m)ⁿ  =  x^mn

Rule 4 : 

(xy)^m  =  x^m ⋅ y^m

Rule 5 : 

(x / y)^m  =  x^m / y^m

Rule 6 : 

x^-m  =  1 / x^m

Rule 7 : 

x⁰  =  1

Rule 8 : 

x¹  =  x

Rule 9 : 

x^m/n  =  y -----> x  =  y^n/m

Rule 10 : 

(x / y)^-m  =  (y / x)^m

Rule 11 : 

a^x  =  a^y -----> x  =  y

Rule 12 : 

x^a  =  y^a -----> x  =  y

Square Root Rules

Rule 1 : 

√a ⋅ √b  =  √(ab)

Rule 2 : 

√a / √b  =  √(a/b)

Rule 3 : 

√a⋅ √a  =  a

Rule 4 : 

√a  =  k ----->  a^1/2  =  k

Rule 5 : 

√a  =  b -----> a  =  b²

Practice Problems

Problem 1 : 

Simplify : 

(a⁷ ⋅ a² ⋅ a^-4) / (a² ⋅ a^-3 ⋅ a⁴)

Solution : 

(a⁷ ⋅ a² ⋅ a^-4) / (a² ⋅ a^-3 ⋅ a⁴)  =  a^7+2-4 / a^2-3+4

(a⁷ ⋅ a² ⋅ a^-4) / (a² ⋅ a^-3 ⋅ a⁴)  =  a⁵ / a³

(a⁷ ⋅ a² ⋅ a^-4) / (a² ⋅ a^-3 ⋅ a⁴)  =  a^5-3

(a⁷ ⋅ a² ⋅ a^-4) / (a² ⋅ a^-3 ⋅ a⁴)  =  a²

Problem 2 : 

Simplify : 

(a⁶ ⋅ b³) / (a² ⋅ b^-3)²

Solution :

(a⁶ ⋅ b³) / (a² ⋅ b^-3)²  =  (a⁶ ⋅ b³) / [(a²)² ⋅ (b^-3)²]

(a⁶ ⋅ b³) / (a² ⋅ b^-3)²  =  (a⁶ ⋅ b³) / (a⁴ ⋅ b^-6)

(a⁶ ⋅ b³) / (a² ⋅ b^-3)²  =  a^6-4 ⋅ b³⁺⁶

(a⁶ ⋅ b³) / (a² ⋅ b^-3)²  =  a²b⁹

Problem 3 : 

If 8^{2n + 3}  =  4^{n + 5}, then find the value of n. 

Solution : 

8^{2n + 3}  =  4^{n + 5}

(2³)^{2n + 3}  =  (2²)^{n + 5}

2^{3(2n + 3)}  =  2^{2(n + 5)}

Equate the exponents.   

3(2n + 3)  =  2(n + 5)

6n + 9  =  2n + 10

4n  =  1

n  =  1/4

Problem 4 :

Simplify the following square root expression :

√40 + √160

Solution : 

Decompose 40 and 160 into prime factors using synthetic division. 

√40  =  √(2 ⋅ 2 ⋅ 2 ⋅ 5)  =  2√10

√160  =  √(2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 5)  =  4√10

So, we have

√40 + √160  =  2√10 + 4√10

√40 + √160  =  6√10

Problem 5 : 

Simplify the following square root expression :

(14√117) ÷ (7√52)

Solution : 

Decompose 117 and 52 into prime factors using synthetic division.

(14√117) ÷ (7√52)  =  14(3√13) ÷ 7(2√13)

(14√117) ÷ (7√52)  =  42√13 ÷ 14√13

(14√117) ÷ (7√52)  =  42√13 / 14√13

(14√117) ÷ (7√52)  =  3

Problem 6 :

Simplify the following square root expression :

(√3)³ + √27  

Solution :

(√3)³ + √27  =  (√3 ⋅ √3 ⋅ √3) + √(3 ⋅ 3 ⋅ 3)

(√3)³ + √27  =  (3 ⋅ √3) + 3√3

(√3)³ + √27  =  3√3 + 3√3

(√3)³ + √27  =  6√3

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