# EXPONENTS AND SQUARE ROOTS

## About "Exponents and Square Roots"

Exponents and Square Roots :

Exponent is a very important part of algebra.

An exponent is just a convenient way of writing repeated multiplications of the same number.

Square root involve the use of the radical sign √.

Sometimes, these are called surds.

If we learn the rules of exponents and square roots, then our enjoyment of learning math will surely increase.

## Exponents and Square Roots

The following stuff are the important components of exponents square roots.

1. Laws of exponents

2. Evaluating expressions with exponents

3. Solving equations with exponents

4. Simplifying square roots

5. Rationalizing the denominator

To know each stuff listed above in detail, please click on it.

## Exponents and Square Roots - Examples

Example 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

Example 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

Example 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

Example 4 :

Solve for k :

5x2 ⋅ 3x4  =  15(xk)2

Solution :

5x2 ⋅ 3x4  =  15(xk)2

(5 ⋅ 3)(x⋅ x4)  =  15x2k

15(x2+4)  =  15x2k

15x6  =  15x2k

Divide each side by 15.

x6  =  x2k

Equate the exponents.

6  =  2k

Divide each side by 2.

3  =  k

So, the value of k is 3.

Example 5 :

If 3x+3 - 3x+2  =  k(3x), then solve for k.

Solution :

3x+3 - 3x+2  =  k(3x)

Using laws of exponents, we have

3x ⋅ 33 - 3x ⋅ 32  =  k(3x)

3x ⋅ 27 - 3x ⋅ 9  =  k(3x)

3x(27 - 9)  =  k(3x)

3x(18)  =  k(3x)

Divide each side by 3x.

18  =  k

So, the value of k is 18.

Example 6 :

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

Example 7 :

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

Example 8 :

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

Example 9 :

If x2y3  =  10 and x3y2  =  8, then find the value of x5y5.

Solution :

x2y3  =  10 -----(1)

x3y2  =  8 -----(2)

Multiply (1) and (2) :

(1) ⋅ (2) -----> (x2y3) ⋅ (x3y2)  =  10 ⋅ 8

x5y5  =  80

So, the value x5yis 80.

Example 10 :

Rationalize the denominator :

(3 - √3) / √3

Solution :

To get rid of the radical in denominator, multiply both numerator and denominator by √3.

(3 - √3) / √3  =  [(3-√3) ⋅ √3] / (√3 ⋅ √3)

(3 - √3) / √3  =  (3√3 - 3) / 3

(3 - √3) / √3  =  3(√3 - 1) / 3

(3 - √3) / √3  =  √3 - 1

After having gone through the stuff given above, we hope that the students would have understood, "Exponents and Square Roots".

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