PRACTICE QUESTIONS OF RADICALS AND EXPONENTS FOR SAT

Problem 1 :

If x^c (3x)² / 9x³ = x⁶ and x ≠ 0, what is the value of c ?

Solution :

x^c (3x)² / 9x³ = x⁶

By distributing the power for all the terms which are multiplied inside the bracket, we get

x^c (9x²) / 9x³ = x⁶

Cancelling the common factors in the numeratror and denominator, we get

x^c / x = x⁶

x^c-1  = x⁶

Since the bases are equal, by equating the powers we get

c - 1 = 6

c = 6 + 1

c = 7

So, the value of c is 7.

Problem 2 :

x^5r/(x^{3r - 2s}) = x^t 

If r + s = 6 and x ≠ 0, what is the value of t in the equation shown ?

a)  6      b)  12    c)  18      d) 30

Solution :

x^5r/(x^{3r - 2s}) = x^t

Using the rule a^m/aⁿ = a^{m - n}

x^{5r - (3r - 2s)} = x^t

x^{5r - 3r + 2s} = x^t

x^{2r + 2s} = x^t

2r + 2s = t

Factoring 2, we get

2(r + s) = t

Applying the value of r + s, we get

2(6) = t

t = 12

So, the value of t is 12. Option b is correct.

Problem 3 :

8 + [√(2x + 29) / 3] = 9

For what value of x is this equation true ?

a)  -10      b)  -2    c)  19      d) No solution

Solution :

8 + [√(2x + 29) / 3] = 9

√(2x + 29) / 3 = 9 - 8

√(2x + 29) / 3 = 1

Multiplying by 3 on both sides, we get

√(2x + 29) = 3

Squaring on both sides, we get

2x + 29 = 3²

2x + 29 = 9

2x = 9 - 29

2x = -20

x = -20/2

x = -10

So, option a is correct.

Problem 4 :

3x = x + 14

√(3z² - 11) + 2x = 22

If z > 0, what is the value of z ?

a)  1      b)  3    c)  5      d) 8

Solution :

3x = x + 14

3x - x = 14

2x = 14

x = 14/2

x = 7

√(3z² - 11) + 2x = 22

Applying the value of x, we get

√(3z² - 11) + 2(7) = 22

√(3z² - 11) + 14 = 22

√(3z² - 11) = 22 - 14

√(3z² - 11) = 8

Squating both sides, we get

(3z² - 11) = 8²

(3z² - 11) = 64

3z² = 64 + 11

3z² = 75

z² = 75/3

z² = 25

z = -5 and 5

Since z > 0, we choose 5.

Problem 5 :

If n³ = -8, what is the value of (n²)³ / (1/n²)

Solution :

n³ = -8

n³ = (-2)³

n = -2

(n²)³ / (1/n²) = ((-2)²)³ / (1/(-2)²)

= (-2)⁶ / (1/4)

= 64 x (4/1)

= 256

Problem 6 :

Which of the following expression is equivalent to -x^1/4 ?

a)  -1/4x      b)  -1/x⁴        c) -∜x       d)  1/∜-x

Solution :

= -x^1/4

Power 1/4 can be written as fourth root.

= -∜x

Problem 7 :

√(3a + 16) - 3 = a - 1

In the equation above, if a ≥ 0, which of the following is possible value of a ?

a)  3      b)  2        c) 1      d)  0

Solution :

√(3a + 16) - 3 = a - 1

√(3a + 16) = a - 1 + 3

√(3a + 16) = a + 2

Squaring both sides, we get

3a + 16 = (a + 2)²

3a + 16 = a² + 4a + 4

a² + 4a + 4 - 3a - 16 = 0

a² + a  - 12 = 0

(a + 4)(a - 3) = 0

a = -4 and a = 3

Based on the condition given, we choose a = 3.

Problem 8 :

√(4n² + 13) - y = 0

If n > 0 and y = 7 in the equation above, what is the value of n ?

a)  2     b)   3     c)  4      d)  5

Solution :

√(4n² + 13) - y = 0

√(4n² + 13) = y

Applying the value of y, we get

√(4n² + 13) = 7

(4n² + 13) = 7²

(4n² + 13) = 49

4n² = 49 - 13

4n² = 36

n² = 36 / 4

n² = 9

n = -3 and n = 3

So, we choose n = 3.

Problem 9 :

If 5x - y = 7, what is the value of 32^x/2^y ?

a)  2⁷     b)   4⁷     c)  8³      d)  The answer cannot be determined

Solution :

5x - y = 7

32^x/2^y = (2⁵)^x/2^y

= 2^5x/2^y

= 2^{5x - y}

Applying the value of 5x - y = 7, we get

= 2⁷

Option a is the answer.

Problem 10 :

If √p + √16 = √81, what is the value of p ?

a)  2     b)   5     c)  9      d)  25

Solution :

√p + √16 = √81

√p  = √81 - √16

Squaring both sides,

p = (√81 - √16)²

p = (9 - 4)²

p = 5²

p = 25

So, option d is correct.