# 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⋅3= 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/y= 210

1/y= (22)5

(1/y)= 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 = 2x, x = ?

Solution :

812 = 2x

8 = 2⋅2⋅2 ==> 23

(23)12 = 2x

236 = 2x

x = 36

Problem 10 :

c2/5 = 4, c = ?

Solution :

c2/5 = 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 :

94x / 273x = ?

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

Solution :

= 94x / 273x

9 = 32 and 27 = 33

= (32)4x / (33)3x

= 38x / 39x

= 38x - 9x

= 3-x

= 1/3x

Problem 12 :

Which of the following is equal to 58x?

I. (54x)4

II. (54x)2

III. (54x)(54x)

(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 (54x)4 = 516x ≠ 58x

II  (54x)2 = 58x

Using the property am x an = am+n, we get

III  (54x)(54x) = 54x+4x = 58x

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 :

 When n = 1n3 ≥ n213 ≥ 12it satisfies When n = 0n3 ≥ n203 ≥ 02it satisfies When n = -1n3 ≥ n2(-1)3 ≥ (-1)2it doesn't satisfy.

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 = b11-7 ==> c4

d-4/ d9 = b-4-9 ==> d-13

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

= a7c4/b4d13

Problem 15 :

274/x = 81, x = ?

Solution :

274/x = 81

Expressing 27 and 81 as a multiple of 3.

33(4/x) = 34

12/x = 4

4x = 12

x = 3

Problem 16 :

3√x-7 = 5, x = ?

Solution :

3√x-7 = 5, x = ?

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

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

= 81/3

= 23(1/3)

= 2

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