# 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 :

xm ⋅ xn  =  xm+n

Rule 2 :

xm ÷ xn  =  xm-n

Rule 3 :

(xm)n  =  xmn

Rule 4 :

(xy)m  =  xm ⋅ ym

Rule 5 :

(x / y)m  =  xm / ym

Rule 6 :

x-m  =  1 / xm

Rule 7 :

x0  =  1

Rule 8 :

x1  =  x

Rule 9 :

xm/n  =  y -----> x  =  yn/m

Rule 10 :

(x / y)-m  =  (y / x)m

Rule 11 :

ax  =  ay -----> x  =  y

Rule 12 :

xa  =  ya -----> 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 ----->  a1/2  =  k

Rule 5 :

√a  =  b -----> a  =  b2

## Practice Problems

Problem 1 :

Simplify :

(a7 ⋅ a⋅ a-4) / (a⋅ a-3 ⋅ a4)

Solution :

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

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

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

(a7 ⋅ a⋅ a-4) / (a⋅ a-3 ⋅ a4)  =  a2

Problem 2 :

Simplify :

(a6 ⋅ b3) / (a⋅ b-3)2

Solution :

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

(a6 ⋅ b3) / (a⋅ b-3) =  (a6 ⋅ b3) / (a4 ⋅ b-6)

(a6 ⋅ b3) / (a⋅ b-3) =  a6-4 ⋅ b3+6

(a6 ⋅ b3) / (a⋅ b-3) =  a2b9

Problem 3 :

If 82n + 3  =  4n + 5, then find the value of n.

Solution :

82n + 3  =  4n + 5

(23)2n + 3  =  (22)n + 5

23(2n + 3)  =  22(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. √117  =  √(3 ⋅ 3 ⋅ 13)√117  =  3√13 √52  =  √(2 ⋅ 2 ⋅ 13)√52  =  2√13

(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 Apart from the stuff given in this section if you need any other stuff in math, please use our google custom search here.

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