USING PROPERTIES OF EXPONENTS PRACTICE PROBLEMS FOR SAT

Problem 1 :

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

(a)  0   (b)  3   (c)  6   (d)  9

Solution :

3^{x - 3} = 27

3^{x - 3} = 3³

Equating the powers, we get

x - 3 = 3

x = 3 + 3

x = 6

Problem 2 :

If 27⁸¹ = 3^x, what is the value of x ?

(a)  27    (b)  84   (c)  100    (d) 243

Solution :

27⁸¹ = 3^x

Decomposing 27, we get 27 = 3³

(3³)⁸¹ = 3^x

3²⁴³ = 3^x

Equating the powers, we get

x = 243

Problem 3 :

If y > 0 and 

what is the value of b ?

(a) -3/2   (b)  -5/2   (c)  3/2    (d)  5/2

Solution :

Problem 4 :

k² x^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 ?

(a)  -1    (b)  0     (c)  1   (d) 2

Solution :

k² x^2a = x^{2a + 2}

k² x^2a = x^2a x² 

Dividing by x^2a on both sides.

k² = x² 

x² - k² = 0

(x + k)(x - k) = 0

x - k = 0

Problem 5 :

x > 1, which of the following could be the value of a - b ?

(a)  3   (b)  4    (c)  5    (d)  6

Solution :

So, option c is correct.

Problem 6 :

If 

where n ≠ 0

Solution :

Problem 7 :

The expression

where h > 0 and k > 0 is equivalent to which of the following ?

Solution :

Problem 8 :

If c^-3d = 1/64 and c and d are positive integers, what is one possible value of d ?

Solution :

c^-3d = 1/64

64 can be written as 4³

c^-3d = 1/4³

c^-3d = 4^-3

(c^d)^-3 = 4^-3

Powers are equal, so equating bases 

c^d = 4

Problem 9 :

Simplify the following :

Solution :

Problem 10 :

If m = 1/√n, where m > 0 and n>0, what is n in terms of m ?

Solution :

m = 1/√n

Take square on both sides.

m² = 1/n

n = 1/m²

Problem 11 :

If c^-1/3 = x, where c > 0 and x > 0, then find the value of c in terms of x.

Solution :

c^-1/3 = x

Moving the power to the other side of the equal sign,

c = x^-3/1

c = x^-3

c = 1/x³

Problem 12 :

If a^-1/2 = 3, what is the value of a ?

(a) -9    (b)  1/9   (c)  1/3   (d) 9

Solution :

a^-1/2 = 3

Moving the power to the other side of the equal sign, we get

a = 3^-2/1

a = 3^-2

a = 1/3²

a = 1/9

Problem 13 :

Let n = 1² + 1⁴ + 1⁶ + 1⁸ + ............ + 1⁵⁰

what is value of n ?

(a)  10   (b)  20   (c)  25   (d) 30

Solution :

n = 1² + 1⁴ + 1⁶ + 1⁸ + ............ + 1⁵⁰

Considering the powers, they are multiples of 2.

n = 1 + 1 + 1 + .......+ 1(25 terms)

n = 25

Problem 14 :

If 4^{2n + 3} = 8^{n + 5}, what is the value of n ?

(a)  6   (b)  7   (c)  8    (d)  9

Solution :

4^{2n + 3} = 8^{n + 5}

Write the bases 4 and 8 as multiples of 2.

(2²)^{2n + 3} = (2³)^{n + 5}

2^{2(2n + 3)} = 2^{3(n + 5)}

Since bases are equal, we can equate the powers.

2(2n + 3) = 3(n + 5)

4n + 6 = 3n + 15

4n - 3n = 15 - 6

n = 9

Problem 15 :

If 2^x/2^y = 2³, then x must equal 

(a) y + 3   (b)  y - 3    (c)  3 - y     (d)  3y

Solution :

2^x/2^y = 2³

2^x 2^-y = 2³

2^x-y = 2³

Equating the powers, we get

x - y = 3

x = 3 + y

So, the answer is y + 3.

Problem 16 :

If x² = y³, for what value of z does x^3z = y⁹ ?

(a) -1   (b)  0    (c)  1     (d)  2

Solution :

x² = y³

Raise power 3 on both sides.

(x²)³ = (y³)³

x⁶ = y⁹  ---(1)

Comparing with x^3z = y⁹ ---(2)

(1) = (2)

 x^3z = x⁶

3z = 6

z = 2

Problem 17 :

If 2^{x + 3} - 2^x = k(2^x) , what is the value of k ?

(a) -1   (b)  0    (c)  1     (d)  2

Solution :

2^{x + 3} - 2^x = k(2^x) 

2^x 2³ - 2^x = k(2^x) 

2^x (2³ - 1) = k(2^x) 

Dividing by 2^x on both sides, we get

 (2³ - 1) = k

k = 7

Problem 18 :

If x^ac x^bc = x³⁰, x > 1 and a + b = 5, what is the value of c ?

(a) 3   (b)  5    (c)  6     (d)  10

Solution :

x^ac x^bc = x³⁰

x^(ac+bc) = x³⁰

Equating the powers, we get

ac + bc = 30

c(a + b) = 30

Here the value of a + b is 5

c(5) = 30

c = 30/5

c = 6