**Exponents questions **are provided here to practice various types of questions based on the concept of exponents and powers. These questions are designed according to the latest CBSE/ICSE syllabus for class 8. Practising these questions will help students prepare for their class tests and even final exams.

Let a be any arbitrary number, now a + a = 2a, a + a + a = 3a, and if we multiply it n times we write a + a + a + … n times = na. So this is about adding a number repeatedly, but if we multiply a number repeatedly like a × a = a^{2}, a × a × a = a^{3}, and a × a × a × a × …n times = a^{n}, then it is called exponents. The number a^{n} is an exponent whose base is a and power is n, and we say it as “a raised to the power n”.

**Law of exponents:**

- a
^{m}× a^{n}= a^{m + n}; Multiplication of exponents with the same base and different power. - a
^{m.}÷ a^{n}= a^{ m – n}; Division of exponents with the same base and different power. - a
^{m}× b^{m}= (ab)^{m}; Multiplication of exponents with the same power and different base. - a
^{m}÷ b^{m}= (a/b)^{m}; Division of exponents with the same power and different base. - a
^{ –m}= 1/a^{m}; Exponents with negative power. - a
^{0}= 1; Exponents with zero power. - (a
^{m})^{n}= a^{ m × n}; Product of powers. - a
^{m/n }=^{n}√a^{m}; Exponents with rational power.

Learn more about exponents and powers.

## Exponents Questions with Solution

Let us solve this practice worksheet on exponents, using the concept and laws of exponents.

**Question 1: Simplify the following:**

**(i) (3/4) ^{8 }× (4/3)^{ 5}_{ } (ii) (5/7)^{5 }× (5/7)^{ –6}**

**Solution:**

(i) (3/4)^{8 }× (4/3)^{ 5}_{ } = (¾)^{8} × (¾)^{ –5}

= (3/4)^{8 – 5} = (3/4)^{ 3}

= 27/48

(ii) (5/7)^{5 }× (5/7)^{ –6} = (5/7)^{5 – 6}

= (5/7)^{ –1} = 7/5

**Question 2: Express each of the following as rational numbers:**

**(i) (4/5) ^{4} (ii) (64/81)^{3/2} (iii) ( –2/5)^{ –4}**

**Solution:**

(i) (4/5)^{4} = 4^{4}/5^{4} = 256/625

(ii) (64/81)^{3/2} = [{64/81}^{1/2}]^{3}

= [8/9]^{3} = 512/729

(iii) ( –2/5)^{ –4} = ( –5/2)^{4}

= 625/16

**Question 3: Simplify each of the following:**

**(i) [(–2/3) ^{ –3} × ( –3/5)^{6}] ÷ (2/3)^{5}**

**(ii) (3/4) ^{ 5} ÷ (81)^{2}**

**× (5/3)**

^{3}

**(iii) [{( –2/3) ^{2 }}^{ –2}]^{ –1}**

**Solution:**

(i) [(–2/3)^{ –3} × ( –3/5)^{6}] ÷ (2/3)^{5}

= [(–3/2)^{ 3} × ( –3/5)^{6}] × (3/2)^{5}

= [3^{3 + 6 + 5}/(2^{3 + 5} × 5^{6})]

= 2^{–8 }3^{14} 5^{ –6}

(ii) (3/4)^{ 5} ÷ (81)^{2} × (5/3)^{3}

= (3/4)^{ 5} ÷ (3^{4})^{2} × (5/3)^{3}

= (3/4)^{ 5} × 3^{ –8} × (5/3)^{3}

= (3^{ 5 –8 –3} × 5^{3})/4^{5}

= 3 ^{–6} 4 ^{–5} 5^{ 3}

(iii) [{( –2/3)^{2 }}^{ –2}]^{ –1}

= ( –2/3)^{ 2 × ( –2) × –1}

= ( –2/3)^{ 4}

= 16/81.

**Question 4: A number when divides ( –15) ^{ –1} results ( –5)^{ –1}. Find the number.**

**Solution:**

Let x be the number such that

( –15)^{ –1} ÷ x = ( –5)^{ –1}

⇒ –1/15 ÷ x = –⅕

⇒ –1/15 × 1/x = –⅕

⇒ –1/15x = –⅕

⇒ 15x = 5

⇒ x = ⅓ or 3^{ –1}

**Question 5: Simplify: **

**\(\begin{array}{l}\frac{10 \times 5^{n + 4}+125 \times 5^{n+2}}{3 \times 5^{n+3}+20 \times 5^{n+1}}\end{array} \)**

**Solution:**

We have

\(\begin{array}{l}\frac{10 \times 5^{n + 4}+125 \times 5^{n+2}}{3 \times 5^{n+3}+20 \times 5^{n+1}}\end{array} \)

\(\begin{array}{l}=\frac{2 \times 5 \times 5^{n + 4}+5^{3} \times 5^{n+2}}{3 \times 5^{n+3}+2^{2}\times 5 \times 5^{n+1}}\end{array} \)

\(\begin{array}{l}=\frac{2 \times 5^{n + 5}+ 5^{n+5}}{3 \times 5^{n+3}+2^{2} \times 5^{n+2}}\end{array} \)

\(\begin{array}{l}=\frac{5^{n + 5}}{5^{n+2}}\left [ \frac{2 + 1}{3 \times 5+2^{2} } \right ]\end{array} \)

\(\begin{array}{l}=5^{n + 5-n-2}\left [ \frac{3}{15+4 } \right ]\end{array} \)

\(\begin{array}{l}=5^{3}\left [ \frac{3}{19 } \right ]=\frac{375}{19}\end{array} \)

\(\begin{array}{l}\therefore \frac{10 \times 5^{n + 4}+125 \times 5^{n+2}}{3 \times 5^{n+3}+20 \times 5^{n+1}}=\frac{375}{19}\end{array} \)

**Question 6: Find the value of p, if 25 ^{(p – 1)} + 100 = 5^{(2p – 1)}.**

**Solution:**

We have 25^{(p – 1)} + 100 = 5^{(2p – 1)}

⇒ 5^{(2p – 1)} – (5^{2})^{(p – 1)} = 100

⇒ 5^{(2p – 1)} – 5^{(2p – 2)} = 100

⇒ 5^{(2p – 2)} × (5 – 1) = 100

⇒ 5^{(2p – 2)} × 4 = 100

⇒ 5^{(2p – 2)} = 25

⇒ 5^{(2p – 2)} = 5^{2}

On comparing the powers,

2p – 2 = 2 ⇒ p = 4

**Question 7: Solve for x: 4 ^{x} + 2^{x} – 20 = 0**

**Solution:**

We have 4^{x} + 2^{x} – 20 = 0

⇒ (2^{2})^{x} + 2^{x} – 20 = 0

⇒ 2^{x} + 2^{x} – 20 = 0

Put 2^{x} = t, the above equation becomes a quadratic equation

t^{2} + t – 20 = 0

⇒ (t + 5)(t – 4) = 0

∴ We get, t = –5 or 4, that is;

2^{x} = – 5_{ }or 2^{x} = 4

Since 2^{x} cannot be negative for any real number x,

2^{x} = 4 = 2^{2}

⇒ x = 2 (on comparing the powers on both sides)

**Question 8: Solve for x: 4 ^{x} – 6(2^{x}) + 8 = 0**

**Solution: **

We have 4^{x} – 6(2^{x}) + 8 = 0

⇒ 2^{2x} – 6(2^{x}) + 8 = 0

Put 2^{x} = t, we get

t^{2} – 6t + 8 = 0

⇒ (t – 4)(t – 2) = 0

∴ We get, t = 2 or 4, that is;

2^{x} = 2 or 2^{x} = 4

⇒ x = 1 or 2 (on comparing the powers on both sides)

**Question 9: Solve of x: (3/5) ^{3} × (3/5)^{ –6} = (3/5)^{ 2x – 1}**

**Solution:**

We have, (3/5)^{3} × (3/5)^{ –6} = (3/5)^{ 2x – 1}

⇒ (3/5)^{3} ^{– 6} = (3/5)^{ 2x – 1}

Comparing powers on both sides,

⇒ 2x – 1 = –3

⇒ x = –1

**Question 10: Simplify: (2 ^{m + 2} – 2^{m})/2^{m} **

**Solution:**

(2^{m + 2} – 2^{m})/2^{m} = 2^{m} (2^{2} – 1)/2^{m}

= 4 – 1 = 3.

## Video Lesson on Exponents

## Related Articles | |

LCM Questions | Integers Questions |

Permutation and Combination Questions | Lines and Angles Questions |

## Practice Questions on Exponents

1. Find: {(3/2)^{ –1} + ( –2/5)^{ –1}}

2. Simplify:

\(\begin{array}{l}\frac{16 \times 2^{n+1}-4\times 2^{2}}{16 \times 2^{n+2}-2 \times 2^{n+2}}\end{array} \)

3. Solve for x:

(i) 7^{x + 1} = 343

(ii) 3^{ –(x + 1)} = 243

(iii) 49^{x} + 1 = 2(7^{x})

4. Simplify: (14^{2} – 13^{2})^{5/3}

5. Find the value of p: 3^{p + 8} = 27^{ 2p + 1}.

Keep visiting BYJU’S to get more such Maths lessons in a simple, concise and easy to understand way. Also, register at BYJU’S – The Learning App to get complete assistance for Maths preparation with video lessons, notes, tips and other study materials.