PROBLEMS WITH NEGATIVE EXPONENTS

Simplify. Your answer should contain only positive exponents.

Problem 1 :

(x^-2 x^-3)⁴

Solution :

Given, (x^-2 x^-3)⁴

= (x^-5)⁴

= x^-20

Converting the negative exponent to positive exponent, we get

= 1/x²⁰

Problem 2 :

(x⁴)^-3 ⋅ 2x⁴

Solution :

Given, (x⁴)^-3 ⋅ 2x⁴

= x^4(-3) ⋅ 2x⁴

= x^-12 ⋅ 2x⁴

= 2 ⋅ x^-12 ⋅ x⁴

= 2 ⋅ x^-8

Sicne we have negative exponent for x, we write the reciprocal of x^-8

= 2 (1/x⁸)

= 2/x⁸

Problem 3 :

(n³)³ ⋅ 2n^-1

Solution :

Given, (n³)³ ⋅ 2n^-1

= n ^{(3 ⋅ 3)} ⋅ 2n^-1

= n⁹ ⋅ 2n^-1

= 2n⁹ ⋅ n^-1

= 2n⁸

Problem 4 :

(2v)²  ⋅ 2v²

Solution :

Given, (2v)²  ⋅ 2v²

= 2² ⋅ v² ⋅ 2v²

= 4v² ⋅ 2v²

= 8v⁴

Problem 5 :

(2x² y  · 4x² y⁴ · 3x)/3x^-3 y²

Solution :

= (2x² y  · 4x² y⁴ · 3x)/3x^-3 y²

Simplifying the numerator :

= 24x²⁺²⁺¹ y⁴⁺¹

= 24x⁵ y⁵

Simplifying the denominator :

= 3x^-3 y²

Converting the negative exponent to positive exponent, we get

= 3 y²/x³

Dividing the numerator by denominator, we get

= 24x⁵ y⁵ / (3 y²/x³)

= 24x⁵ y⁵  · (x³ / 3 y²)

= (24/3) x⁵⁺³ y^5-2

= 8 x⁸ y³

Problem 6 :

(2 y³· 3 x y³) / 3 x² y⁴

Solution :

= (2 y³· 3 x y³) / 3 x² y⁴

= (6 y³⁺³ x) / 3 x² y⁴

= (6/3) y⁶ x^1-2 y⁴

= 2 y⁶⁺⁴ x^-1

= 2 y¹⁰ / x¹

= 2 y¹⁰ / x

Problem 7 :

(x³· y³· x³) / 4 x²

Solution :

= (x³· y³· x³) / 4 x²

= (x³⁺³· y³) / 4 x²

= (x⁶  y³) / 4 x²

= (x^{6 - 2}  y³) / 4

= (x⁴  y³) / 4

Problem 8 :

(3 x²· y²) / (2 x^-1 · 4 yx²)

Solution :

= (3 x²· y²) / (2 x^-1 · 4 yx²)

= (3 x²· y²) / (8 x^-1+2 y)

= (3 x²· y²) / (8 x¹ y)

= (3/8) x^{2 - 1}· y^{2 - 1}

= (3/8) x y

= 3xy / 8

Problem 9 :

x / (2 x⁰)²

Solution :

= x / (2 x⁰)²

= x / (2 (1))²

= x / 2²

= x/4

Problem 10 :

2m^-4 / (2m^-4)³

Solution :

= 2m^-4 / (2m^-4)³

= 2m^-4 / 2³(m^-4)³

= 2m^-4 / 8m^-12

= (2/8) m^-4+12

= (2/8) m⁸

= (1/4) m⁸

Problem 11 :

(2m²)^-1 / m²

Solution :

= (2m²)^-1 / m²

= 1 / (2m²)¹

= 1 / 2m²

Problem 12 :

(2x³)/(x^-1)³

Solution :

= (2x³)/(x^-1)³

= (2x³)/x^-3

= 2x³ · x³

= 2x³⁺³

= 2x⁶

Problem 13 :

(a^-3 b^-3)⁰

Solution :

Given, (a^-3 b^-3)⁰

= (a^-3)⁰ ⋅ (b^-3)⁰

= 1 ⋅  1

= 1

Problem 14 :

x⁴ y³ ⋅ (2y²)⁰

Solution :

Given,  x⁴ y³ ⋅ (2y²)⁰

=  x⁴ y³  ⋅ 1

= x⁴ y³

Problem 15 :

ba⁴ ⋅ (2ba⁴)^-3

Solution :

Given, ba⁴ ⋅ (2ba⁴)^-3

= ba⁴ ⋅ 2^-3 ⋅ b^-3 ⋅ (a⁴)^-3

= ba⁴ ⋅ 2^-3 ⋅ b^-3 ⋅ a^-12 

= b^-2  ⋅ 2^-3 ⋅ a^-8

= 1/(b²  ⋅ 2³ ⋅ a⁸)

= 1/8a⁸ b²

Problem 16 :

(2x⁰y²)^-3 ⋅ 2yx³

Solution :

Given, (2x⁰y²)^-3 ⋅ 2yx³

= (2y²)^-3 ⋅ 2yx³

= 2yx³ / (2y²)³

= 2yx³ / 8y⁶

= 2x³ / 8y^{6 - 1}

= (2/8) x³ / y⁵

Problem 17 :

Express (15³)^-16 as single exponent of 15.

Solution :

= (15³)^-16

Since we have power raised by another power, we have to multiply the powers.

= 15^-48

Converting the negative exponent as positive exponent, we get

= 1/15⁴⁸

Problem 18 :

By what number should (7^-2) be multiplied so that the product may be equal to (343)^-1 ?

Solution :

= (7^-2)

Let x be the required number to be multiplied.

(7^-2) ⋅ x = (343)^-1

x = (343)^-1 / (7^-2)

x = (1/343) / (1/7²)

= (1/343) / (1/49)

= 49/343

= 1/7

so, the required number is 1/7.

Problem 19 :

By what number should (-3/4)⁵ be multiplied so that the product may be equal to (-64/27)^-1 ?

Solution :

= (-3/4)⁵ 

Let x be the required number to be multiplied.

(-3/4)⁵  ⋅ x = (-64/27)^-1

x = (-64/27)^-1 / (-3/4)⁵

x = (-27/64) / (-3/4)⁵

x = (-27/64) / (-81/1024)

= (27/64) ⋅ (1024/81)

= 16/3

So, the required number is 16/3.

Problem 20 :

By what number should (-512)^-1 be divided so that the quotient may be equal to 

8^-2 ?

Solution :

= (-512)^-1 

Let x be the required number.

(-512)^-1 / x = 8^-2 

x = (-512)^-1 / 8^-2 

Converting the negative exponent to positive exponent, we get

x = 8² / (-512)

= -64/512

= -1/8

So, the required number is -1/8

Problem 21 :

By what number should (144/225)^-1 be divided so that the quotient may be equal to 

(12/15)^-4 ?

Solution :

= (144/225)^-1 

Let x be the required number.

(144/225)^-1 / x = (12/15)^-4 

x = (144/225)^-1 /  (12/15)^-4 

Converting the negative exponent to positive exponent, we get

x = (225/144)¹ /  (15/12)⁴ 

= (225/144) / (15/12)⁴ 

= (225/144) ⋅ (12⋅12⋅12⋅12)/(15⋅15⋅15⋅15)

= 144/225

So, the required number is 144/225.

Problem 22 :

Simplify 6^-2 + (3/2)^-2

Solution :

= 6^-2 + (3/2)^-2

Converting the negative exponent to positive exponent by writing it as reciprocal, we get

= 1/6² + (2/3)²

= 1/36 + (4/9)

LCM of 9 and 36 = 36

= 1/36 + 16/36

= (1 + 16)/36

= 17/36

So, the simplified value is 17/36

Problem 23 :

Find the value of x if [(3/7)³]^-2 = (3/7)^2x

Solution :

[(3/7)³]^-2 = (3/7)^2x

(3/7)^-6 = (3/7)^2x

-6 = 2x

x = -6/2

x = -3

So, the value of x is -3.