SAT MATH QUESTIONS ON EXPONENTS AND RADICALS

Question 1 :

If 3^{x - 3} = 27, what is the value of x?

Answer :

In this question, we are trying to solve for the variable in exponent. In an equation, the idea to solve for the variable in exponent is to get the same base on both sides of the equation.

We have 27 on the right side of the given equation.

27 is a power of 3, that is 27 = 3³.

3^{x - 3} = 27

3^{x - 3} = 3³

In the equation, we have the same base on both sides. So, we can equate the exponents.

x - 3 = 3

Add 3 to both sides.

x = 6

Question 2 :

If 9^x = 27⁸², what is the value of x?

Answer :

9^x = 27⁸²

(3²)^x = (3³)⁸² 

3^2x = 3²⁴⁶

2x = 246

Divide both sides by 2.

x = 123

Question 3 :

If y⁸ = m and y⁹ = 2/3, what is the value of y in terms of m?

Answer :

y⁸ = m ----(1)

y⁹ = 2/3 

----(2)

Divide (2) by (1).

(2) ÷ (1) ----> y^m ÷ yⁿ = m ÷ (2/3)

y^{9 - 8} = m ⋅ (3/2)

y = 3m/2

Question 4 :

k²x^2a = x^{2a + 2}

In the equation above, k, x and a are positive integers greater than 1. What is the value of x - k?

Answer :

k²x^2a = x^{2a + 2}

k²x^2a = x^2ax²

Divide both sides by x^2a.

k² = x²

Since both k and x are positive,

k = x

Therefore,

x - k = 0

Question 5 :

In the equation above, x > 1. Find the value of (a - b) such that (a - b) > 0.

Answer :

(a + b)² = 25

Taking square root on both sides.

√(a + b)² = √25

a + b = ±5

a + b = 5  or  a + b = -5

Since (a + b) > 0,

a + b = 5

Question 6 :

Which of the following is equal to ³√x² ⋅ √x³ ?

A) x     B) x^9/4     C) x^13/6        D) x³

Answer :

³√x² ⋅ √x³ = (x²)^1/3 ⋅ (x³)^1/2

= x^2/3 ⋅ x^3/2

= x^{2/3 + 3/2}

= x^4/6 + 9/6

= x^{(4 + 9)/6}

= x^13/6

The correct answer choice is (C).

Question 7 :

If 3√x³ = √72, what is the value of x?

Answer :

3√x³ = √72

√9√x³ = √72

√(9x³) = √72

Squaring both sides,

9x³ = 72

Divide both sides by 9.

x³ = 8

x³ = 2³

x = 2

Question 8 :

If 3x - y = 12, what is the value of 8^x ÷ 2^y?

Answer :

8^x ÷ 2^y = (2³)^x ÷ 2^y

= 2^3x ÷ 2^y

= 2^{3x - y}

Substitute 3x - y = 12.

= 2¹²

Question 9 :

In the equation above, if x > 1 and a + b = 2, what is the value of (a - b)?

Answer :

a² - b² = 16

(a + b)(a - b) = 16

Substitute a + b = 2.

2(a - b) = 16

Divide both sides by 2.

a - b = 8

Question 10 :

In the expression above x > 1 and y > 1. Which of the following is equivalent to the above expression ?

Answer :

The correct answer choice is (D).

Question 11 :

The perimeter of an equilateral triangle is 624 centimeters. The height of this triangle is k√3 centimeters, where k is a constant. What is the value of k?

Answer :

Perimeter of the equilateral triangle = 624

Let x be the side length of equilateral triangle.

3x = 624

x = 624/3

x = 208

Area of any triangle = (1/2) x base x height

Area of equilateral triangle = (√3/4) x²

= (1/2) (1/2) (√3) 208²

= (1/2) x 208 x (√3) 104

It is clear base of the equilateral triangle is 208. Then the value of k is 104.

Question 12 :

√(x - 2)² = √(3x + 34)

What is the smallest possible solution for x ?

Answer :

√(x - 2)² = √(3x + 34)

Canceling square and radical, we get

(x - 2) = √(3x + 34)

Squaring on both sides

(x - 2)² = 3x + 34

x² - 2(x) (2) + 2² = 3x + 34

x² - 4x + 4 = 3x + 34

x² - 4x - 3x + 4 - 34 = 0

x² - 7x - 30 = 0

x² - 10x + 3x - 30 = 0

x(x - 10) + 3(x - 10) = 0= 0

(x - 10)(x + 3) = 0

x = 10 and x = -3

So, the smallest possible value of x is -3.

Question 13 :

14x/7y = 2√(w + 19)

The given equation relates the distinct positive real numbers w, x, and y. Which equation correctly expresses w in terms of x and y ?

A)  w = √(x/y) - 19       B)  w = √(28x/14y) - 19

C)  w = (x/y)² - 19          D)  w = (28x/14y)² - 19

Answer :

14x/7y = 2√(w + 19)

Dividing by 2, we get

14x/14y = √(w + 19)

x/y = √(w + 19)

Squaring on both sides

w + 19 = (x/y)²

w = (x/y)² - 19

So, option C is correct.

Question 14 :

A right triangle has sides of length 2√2, 6√2 and √80 units. What is the area of triangle, in square units ?

A)  8√2 + √80       B)  12    C) 24√80         D)  24

Answer :

Since these are the sides of the triangle, it must satisfy Pythagorean theorem. 

 (√80)² = (2√2)² + (6√2)²

80 = 4(2) + 36(2)

80 = 8 + 72

80 = 80

So, the base of the triangle is 2√2 and height is 6√2

Area of triangle = (1/2) x base x height

= (1/2) x 2√2 x 6√2

= 6(2)

= 12 square units

So, the area of the triangle is 12 square units.

Question 15 :

The expressions 6(3⁵ x⁴⁵)^1/5 ⋅ (2⁸ x)^1/8 is equivalent to ax^b,  where a and b are positive constants and x > 1, what is the value of a + b ? 

Answer :

= 6(3⁵ x⁴⁵)^1/5 ⋅ (2⁸ x)^1/8

= 6(3⁵ x^{9 (5)})^1/5 ⋅ (2⁸ x)^1/8

= 6[(3x⁹)⁵]^1/5 ⋅ (2⁸)^1/8 x^1/8

= 6(3x⁹) ⋅ 2 x^1/8

= 12(3x⁹x^1/8)

= 36 x^9+1/8

= 36 x^73/8

Comparing with ax^b

a = 36 and b = 73/8

a + b = 36 + (73/8)

 = (288 + 73)/8

= 361/8

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