EXPONENTS AND SQUARE ROOTS WORKSHEET

Problem 1 : 

Simplify : 

(a⁷ ⋅ a² ⋅ a^-4) / (a² ⋅ a^-3 ⋅ a⁴)

Problem 2 : 

Simplify : 

(a⁶ ⋅ b³) / (a² ⋅ b^-3)²

Problem 3 :

If a^-1/2  =  5, then find the value of a. 

Problem 4 : 

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

Problem 5 :

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

Problem 6 :

Simplify the following square root expression :

√40 + √160

Problem 7 : 

Simplify the following square root expression :

√27 ⋅ √3

Problem 8 : 

Simplify the following square root expression :

(14√117) ÷ (7√52)

Problem 9 :

Simplify the following square root expression :

(√3)³ + √27

Problem 10 :

Rationalize the denominator :  

(3 - √3) / √3

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 a^-1/2  =  5, then find the value of a. 

Solution : 

a^-1/2  =  5

a  =  5^-2/1

a  =  5^-2

a  =  1/5²

a  =  1/25

Problem 4 : 

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

Solution : 

3^x+3 - 3^x+2  =  k(3^x)

Using laws of exponents, we have

3^x ⋅ 3³ - 3^x ⋅ 3²  =  k(3^x)

3^x ⋅ 27 - 3^x ⋅ 9  =  k(3^x)

3^x(27 - 9)  =  k(3^x)

3^x(18)  =  k(3^x)

Divide each side by 3^x.

18  =  k

So, the value of k is 18.

Problem 5 :

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 x⁵y⁵ is 80.

Problem 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

Problem 7 : 

Simplify the following square root expression :

√27 ⋅ √3

Solution : 

√27 ⋅ √3  =  √(27 ⋅ 3)

√27 ⋅ √3  =  √(3 ⋅ 3 ⋅ 3 ⋅ 3)

√27 ⋅ √3  =  3 ⋅ 3

√27 ⋅ √3  =  9

Problem 8 : 

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 9 :

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

Problem 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 

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