SAT MATH QUESTIONS ON RADICALS AND POWERS

Problem 1 :

9³ x 27² = 3ⁿ, n = ?

Solution :

9³ x 27² = 3ⁿ, n = ?

By writing 9 and 27 in exponential form, we get

9 = 3 ⋅ 3 ==> 3²

27 = 3 ⋅ 3 ⋅ 3 ==> 3³

(3²)³ x (3³)² = 3ⁿ

Power raised to another power. So, we multiply the powers.

3⁶⋅3⁶ = 3ⁿ

Bases are same, so use one base and add powers.

3⁶⁺⁶ = 3ⁿ

On both side of the equal sign bases are same. So, equate the powers.

3¹² = 3ⁿ

Problem 2 :

4^-a = 64, a = ?

Solution :

4^-a = 64

We can decompose 64 as a multiple of 4. So, we get

4^-a = 4³

Both sides bases are same. So, equating powers, we get

a = -3

Problem 3 :

x^-3 = 1/8, x = ?

Solution :

Given :

x^-3 = 1/8, x = ?

1/x³ = 1/8

1/x³ = 1/2³

(1/x)³ = (1/2)³

Since the bases are equal, we can equate the bases.

1/x = 1/2

Problem 4 :

5√8+7√32 = 

(a) 18√2  (b) 38√2  (c) 23√4  (d) 33√3  (e)  38√3

Solution :

5√8+7√32

By simplifying √8 and √32, we get

√8 = √(2⋅2⋅2) and √32 = √(2⋅2⋅2⋅2⋅2)

√8 = 2√2 and √32 = 4√2

= 5(2√2) + 7(4√2)

= 10√2 + 28√2

= 38√2

Problem 5 :

4√18 x 11√12 = 

(a) 12√6  (b) 34√6  (c) 264√6  (d) 264√3  (e)  264√2

Solution :

4√18 x 11√12 = 

√18 = √(2 ⋅ 3 ⋅ 3) ==> 3√2

√12 = √(2 ⋅ 2 ⋅ 3)  ==> 2√3

4√18 x 11√12 = 4(3√2) x 11(2√3)

= 12√2 x 22√3

= 264√6

Problem 6 :

y^-5 = 1024, y = ?

Solution :

y^-5 = 1024

1/y⁵ = 2¹⁰

1/y⁵ = (2²)⁵

(1/y)⁵ = 4⁵

1/y = 4

y = 1/4

Problem 7 :

2^-n = 1/256, n = ?

Solution :

2^-n = 1/256

2^-n = 1/2⁸

256 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2

256 = 2⁸

1/2ⁿ = 1/2⁸

(1/2)ⁿ = (1/2)⁸

So, the value of n is 8.

Problem 8 :

√x = 4a²bc³, x = ?

Solution :

Given :

√x = 4a²bc³

To remove the square root, take squares on both sides.

x = (4a²bc³)²

x = 4² a⁴b²c⁶

x = 16 a⁴b²c⁶

Problem 9 :

8¹² = 2^x, x = ?

Solution :

8¹² = 2^x

8 = 2⋅2⋅2 ==> 2³

(2³)¹² = 2^x

2³⁶ = 2^x

x = 36

Problem 10 :

c^2/5 = 4, c = ?

Solution :

c^2/5 = 4

Raise power 5 on both sides.

c² = 4⁵

Take square root on both sides.

c = √4⁵

c = √(4⋅4⋅4⋅4⋅4)

c = 4⋅4√4

c = 16√(2⋅2)

c = 16⋅2

c = 32

Problem 11 :

9^4x / 27^3x = ?

(a) 9^x  (b) 1/3^x  (c) 1/9x  (d) 3^x  (e)  1/4^x

Solution :

= 9^4x / 27^3x

9 = 3² and 27 = 3³

= (3²)^4x / (3³)^3x

= 3^8x / 3^9x

= 3^{8x - 9x}

= 3^-x

= 1/3^x

So, the answer is 1/3^x.

Problem 12 :

Which of the following is equal to 5^8x?

I. (5^4x)⁴

II. (5^4x)²

III. (5^4x)(5^4x)

(a) I  (b) II    (c) III  (d)  I and II    (e)  II and III

Solution :

We have power raised to another power. So, we will multiply the powers.

I (5^4x)⁴ = 5^16x ≠ 5^8x

II  (5^4x)² = 5^8x

Using the property a^m x aⁿ = a^m+n, we get

III  (5^4x)(5^4x) = 5^4x+4x = 5^8x

So, II and III are correct.

Problem 13 :

n³ ≥ n² for which of the following?

I. n = 1

II. n = 0

III. n = −1

A. I   B. II   C. III   D. I and II   E. II and III

Solution :

So, I and II are correct.

Problem 14 :

Simplify (a²b^-6c¹¹d^-4) / (a^-5b^-2c⁷d⁹)

Solution :

(a²b^-6c¹¹d^-4) / (a^-5b^-2c⁷d⁹)

a²/a^-5 = a²⁺⁵ ==> a⁷

b^-6/ b^-2 = b^-6+2 ==> b^-4

c¹¹/ c⁷ = b^11-7 ==> c⁴

d^-4/ d⁹ = b^-4-9 ==> d^-13

(a²b^-6c¹¹d^-4) / (a^-5b^-2c⁷d⁹) = a⁷b^-4c⁴d^-13

= a⁷c⁴/b⁴d¹³

Problem 15 :

27^4/x = 81, x = ?

Solution :

27^4/x = 81

Expressing 27 and 81 as a multiple of 3.

3^3(4/x) = 3⁴

12/x = 4

4x = 12

x = 3

Problem 16 :

3√x-7 = 5, x = ?

Solution :

3√x-7 = 5, x = ?

Add 7 on both sides.

3√x = 5+7

3√x = 12

Divide by 3 on both sides.

√x = 4

Take squares on both sides

x = 4²

x = 16

Problem 17 :

9√x - 7√x - 36 = -16, x = ?

(a) 5   (b) 10   (c) 20   (d) 50  (e) 100

Solution :

9√x - 7√x - 36 = -16, x = ?

2√x - 36 = -16

Add 36 on both sides.

2√x  = -16+36

2√x  = 20

Divide by 2 on both sides.

√x  = 10

Take squares on both sides.

x  = 10²

x = 100

Problem 18 :

x = 2, y = x², (y²-x³)(x²/3y) = 

Solution :

= (y²-x³)^{(x^2/3y)}

By applying the value of x and y given above.

x = 2, y = 2² ==> 4

= (4²-2³)^(4/3(4))

= (16-8)^1/3

= 8^1/3

= 2^3(1/3)

= 2