Exponents and Powers Class 7 Questions

Exponents and Powers Class 7 questions and answers may aid students in quickly understanding the concept. Exponents and Powers Class 7 gives an idea about the difference between the exponents and powers and it is the basis of higher-level ideas. These questions might help students gain a quick overview of the topics and practice answering them to improve their understanding. Learn the entire explanations for each question to double-check your answers. Exponents and Powers Class 7 can be found here.

Exponents and Powers Class 7 Questions with Solutions

1. How do you write 10,000 using exponents in a concise form?

Solution:

The number 10⁴ can be expressed as the number 10000.

The product 10×10×10×10 is represented by the abbreviated notation 10⁴. The base here is ’10,’ and the exponent is ‘4.’ The number 10⁴ is read as 10 to the fourth power, or simply 10 to the power of 4. Therefore, the exponential form of 10,000 is denoted as 10⁴.

2. As a product of powers of prime factors, write the following numbers:

(a) 72 (b) 432 (c) 1000

Solution:

(a) Given number: 72

The number 72 can be written as the product of 2 and 36.

I.e.,72 = 2 × 36

Further, 36 is written as the product of 2 and 18.

72 = = 2 × 2 × 18

Again, 18 is written as the product of 2 and 9.

72 = 2 × 2 × 2 × 9

And we know that, the product of 3 and 3 is 9

72 = 2 × 2 × 2 × 3 × 3

Hence, in exponential form, 72 is written as follows:

72 = 2³ × 3²

Therefore, 72 is written as the product of powers of prime factors is 2³ × 3².

(b) Given number: 432

The number 432 written in the product form is:

432 = 2 × 216 = 2 × 2 × 108

432 = 2 × 2 × 2 × 54

432 = 2 × 2 × 2 × 2 × 27

432 = 2 × 2 × 2 × 2 × 3 × 9

432 = 2 × 2 × 2 × 2 × 3 × 3 × 3

Hence, 432 written in product of powers of prime factors is:

432 = 2⁴ × 3³.

(c) Given number: 1000

As we know, 1000 is the product of 2 and 500.

1000 = 2 × 500

Again 500 is written as the product of 2 and 250.

1000 = 2 × 2 × 250

Further, 250 is the product of 2 and 125.

1000 = 2 × 2 × 2 × 125

Also, the product of 25 and 5 is 125.

1000 = 2 × 2 × 2 × 5 × 25

Further, 5 multiplied by 5 is 25.

1000 = 2 × 2 × 2 × 5 × 5 × 5

Thus,1000 = 2³ × 5³, which is the product of powers of prime factors.

3. Check whether which one is greater (5²) × 3 or (5²)³?

Solution:

To find: Whether (5²) × 3 is greater or (5²)³ is greater

As we know (5²) × 3 means 5² is multiplied by 3

i.e., 5 × 5 × 3 = 75.

Similarly, (5²)³ means 5² is multiplied by itself three times.

It means, 5²× 5² × 5² = 5⁶.

Thus, 5 to the power of 6 equals 15,625.

I.e., (5²)³ = 15,625

Hence, we can say (5²)³ is larger/greater than (5²) × 3.

4. Convert the following terms into the exponential form:

(a) (4 × 5)⁵ (b) (- 2p)³

Solution:

(a) Given expression: (4 × 5)⁵

Now the given expression can be written in the form:

= (4 × 5) × (4 × 5) × (4 × 5) × (4 × 5) × (4 × 5)

= (4 × 4 × 4 × 4 × 4) × (5 × 5× 5 × 5 × 5)

= 4⁵ × 5⁵

(b) Given expression: (- 2p)³

= (- 2 × p)³

= (- 2 × p) × (– 2 × p) × (– 2 × p)

= (- 2) × (– 2) × (– 2) × (p × p × p)

= (- 2)³ × (p)³

5. Determine the exponential form for the given expression: 8 × 8 × 8 × 8 taking base as 2.

Solution:

We know that 8 × 8 × 8 × 8 can be written as 8⁴.

Also, 8 can be written as 2 × 2 × 2

Hence, 8 = 2³

Therefore 8⁴ equals (2³)⁴

(2³)⁴ = 2³× 2³× 2³× 2³

Using the property, (a^m)ⁿ = a^m, we can write

(2³)⁴ = 2^{3 × 4}

= 2¹².

Also, read: Laws of Exponents.

6. Simplify the expression: 2³ × p³ × 5p⁴

Solution:

2³ × p³ × 5p⁴ = 2³ × p³ × 5 × p⁴

= 2³ × 5 × p³ × p⁴

= 8 × 5 × p^{3 + 4}

= 40 p⁷

Hence, 2³ × p³ × 5p⁴ = 40 p⁷.

7. Convert the given number into standard form:

(i) 4323.3 (ii) 26950 (iii) 3,630,000

Solution:

(i) 4323.3 = 4.3233 × 1000 = 4.3233 × 10³

(ii) 26950 = 2.695 × 10,000 = 2.695 × 10⁴

(iii) 3,630,000 = 3.63 × 1,000,000 = 3.63 × 10⁶

8. Check whether the given expression is true or false and justify your answer:

(a) 10 × 10¹¹ = 100¹¹

(b) 3³ > 5²

(c) 1³× 3² = 4⁵

Solution:

(a) 10 × 10¹¹ = 100¹¹

First take the LHS of the equation:

LHS = 10 × 10¹¹

Using the laws of exponents we can write:

= 10^{11 + 1} [As, a^m × aⁿ = a^{m + n}]

= 10¹².

Now, take the RHS of the equation:

RHS = 100¹¹

= (10² )¹¹

Since, (a^m)ⁿ = a^mn

= 10²²

Hence,10¹² is not equal to 10²²

Therefore, the given statement is false.

(b) 3³ > 5²

LHS = 3³ = 3 × 3 × 3 = 27

RHS = 5² = 5× 5 = 25

Therefore, 2³ > 5²

Hence, the given statement is true.

(c) 1³ × 3² = 4⁵

LHS = 1³ × 3²

= 1 × 1 × 1 × 3 × 3 = 9

RHS = 4⁵

= 4 × 4 × 4 × 4 × 4 = 1024

Therefore, the given equation 1³ × 3² = 4⁵ is false.

9. Show the expression 108 × 192 as a product of prime factors in exponential form.

Solution:

Given expression: 108 × 192 …(1)

The prime factorisation of 108 is 2 × 2 × 3 × 3 × 3

108 = 2² × 3³ …(2)

The prime factorisation of 192 is 2 × 2 × 2 × 2 × 2 × 2 × 3

192 = 2⁶ × 3 …(3)

Now, substitute (2) and (3) in (1),

108 × 192 is written as:

= (2² × 3³) × (2⁶ × 3)

Now, using the property, a^m × aⁿ = a^{m + n},

= 2^{2 + 6} × 3^{3 + 1}

= 2⁸ × 3⁴

Therefore,108 × 192 in exponential form is 2⁸ × 3⁴.

10. Convert the number appearing on the given statement in standard form:

“There are an average of 100,000,000,000 stars in the galaxy”.

Solution:

Given statement: There are an average of 100,000,000,000 stars in the galaxy

Thus, the given statement is expressed as follows:

There are an average of 1 × 10¹¹ stars in the galaxy.

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Practice Questions

1. Write the number 70,00,000 in standard form.

2. Write the number in expanded form 104278.

3. Simplify and write the answer in the exponential form: (6² × 6⁴) ÷ 6³.

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