Exponents and Powers | Grade 7th - Mathematics Chapter Summary & NCERT CBSE Study Resources

Study Materials

7th

7th - Mathematics

Exponents and Powers

Overview of the Chapter: Exponents and Powers

This chapter introduces students to the concept of exponents and powers, which are fundamental in understanding large numbers and simplifying mathematical expressions. The chapter covers the laws of exponents, standard forms, and practical applications of exponents in real-life situations.

Key Concepts

Exponents: An exponent refers to the number of times a number is multiplied by itself. For example, in 5³, 5 is the base and 3 is the exponent.

Powers: The value obtained when a number is raised to an exponent is called its power. For example, 5³ = 125, where 125 is the power.

Laws of Exponents

Product of Powers: a^m × aⁿ = a^m+n
Quotient of Powers: a^m ÷ aⁿ = a^m-n (where a ≠ 0)
Power of a Power: (a^m)ⁿ = a^m×n
Power of a Product: (a × b)^m = a^m × b^m
Power of a Quotient: (a ÷ b)^m = a^m ÷ b^m (where b ≠ 0)

Standard Form

Numbers can be expressed in standard form using exponents, especially for very large or very small numbers. For example, 3000 can be written as 3 × 10³.

Applications of Exponents

Exponents are used in various real-life applications such as scientific notation, computing areas and volumes, and in financial calculations like compound interest.

Solved Examples

Simplify: 2³ × 2⁴ = 2³⁺⁴ = 2⁷ = 128
Express 4500 in standard form: 4.5 × 10³

Practice Questions

Evaluate: 5² × 5³
Simplify: (3²)⁴
Express 0.00045 in standard form.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:

Express 8 as a power of 2.

Answer:

2³

Question 2:

What is the value of 5⁰?

Answer:

Question 3:

Write 10,000 in exponential form.

Answer:

10⁴

Question 4:

Simplify 3² × 3³.

Answer:

3⁵

Question 5:

Find the value of (2³)².

Answer:

2⁶

Question 6:

Express 1,000,000 as a power of 10.

Answer:

10⁶

Question 7:

What is the standard form of 7 × 10³ + 2 × 10¹?

Answer:

7020

Question 8:

Evaluate 4³ ÷ 4².

Answer:

Question 9:

Write 0.001 as a power of 10.

Answer:

10^-3

Question 10:

Simplify (5²)⁰.

Answer:

Question 11:

Express 81 as a power of 3.

Answer:

3⁴

Question 12:

What is the value of 6¹?

Answer:

Question 13:

Express 125 as a power of 5.

Answer:

125 can be expressed as 5³ because
5 × 5 × 5 = 125.

Question 14:

What is the value of exponent in 7⁴?

Answer:

The exponent in 7⁴ is 4, which tells how many times the base (7) is multiplied by itself.

Question 15:

Simplify: 2³ × 2⁵.

Answer:

Using the law of exponents for multiplication,
2³ × 2⁵ = 2³⁺⁵ = 2⁸ = 256.

Question 16:

Write 10,000 in exponential form.

Answer:

10,000 can be written as 10⁴ because
10 × 10 × 10 × 10 = 10,000.

Question 17:

What is the value of 3⁰?

Answer:

Any non-zero number raised to the power of 0 is 1.
So, 3⁰ = 1.

Question 18:

Express 1,000,000 as a power of 10.

Answer:

1,000,000 can be expressed as 10⁶ because
10 × 10 × 10 × 10 × 10 × 10 = 1,000,000.

Question 19:

Simplify: (5²)³.

Answer:

Using the power of a power rule,
(5²)³ = 5^2×3 = 5⁶ = 15,625.

Question 20:

What is the base in the expression 9³?

Answer:

The base in 9³ is 9, which is the number being multiplied by itself.

Question 21:

Find the value of 4³ ÷ 4².

Answer:

Using the law of exponents for division,
4³ ÷ 4² = 4^3-2 = 4¹ = 4.

Question 22:

Write 0.001 in exponential form with base 10.

Answer:

0.001 can be written as 10^-3 because
1 ÷ (10 × 10 × 10) = 0.001.

Question 23:

What is the value of 2⁰?

Answer:

Any non-zero number raised to the power 0 is 1.
So, 2⁰ = 1.

Question 24:

Simplify: 3² × 3³.

Answer:

Using the law of exponents, a^m × aⁿ = a^m+n.
So, 3²⁺³ = 3⁵ = 243.

Question 25:

Find the value of (10⁴ ÷ 10²).

Answer:

Using the law of exponents, a^m ÷ aⁿ = a^m-n.
So, 10^4-2 = 10² = 100.

Question 26:

Write the expanded form of 7.5 × 10³.

Answer:

The expanded form is 7.5 × 1000 = 7500.
This is because 10³ = 1000.

Question 27:

What is the standard form of 0.00045 using exponents?

Answer:

0.00045 can be written as 4.5 × 10^-4.
The decimal is moved 4 places to the right.

Question 28:

Evaluate: (2³)².

Answer:

Using the power of a power rule, (a^m)ⁿ = a^m×n.
So, 2^3×2 = 2⁶ = 64.

Question 29:

If 5^x = 125, find the value of x.

Answer:

125 is 5³.
So, 5^x = 5³.
Thus, x = 3.

Question 30:

Simplify: 10⁵ ÷ 10⁵.

Answer:

Using the law of exponents, a^m ÷ aⁿ = a^m-n.
So, 10^5-5 = 10⁰ = 1.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:

What is the value of (3⁰ + 5⁰)?

Answer:

Any non-zero number raised to the power 0 is 1.
So, 3⁰ = 1 and 5⁰ = 1.
Thus, 1 + 1 = 2.

Question 2:

Simplify: (2³ × 2⁴) ÷ 2⁵.

Answer:

Using the laws of exponents:
2³ × 2⁴ = 2³⁺⁴ = 2⁷
Now, 2⁷ ÷ 2⁵ = 2^7-5 = 2² = 4.

Question 3:

Write 0.0001 in exponential form.

Answer:

0.0001 can be written as 10^-4 because:
10^-4 = 1/10⁴ = 1/10000 = 0.0001.

Question 4:

Find the value of (10² × 10³ × 10⁴).

Answer:

Using the law of exponents for multiplication:
10² × 10³ × 10⁴ = 10²⁺³⁺⁴ = 10⁹ = 1,000,000,000.

Question 5:

Express 8 × 8 × 8 × 8 using exponents.

Answer:

8 multiplied four times can be written as 8⁴.

Question 6:

What is the standard form of 5,670,000?

Answer:

The standard form of 5,670,000 is 5.67 × 10⁶.
We move the decimal 6 places to the left.

Question 7:

Simplify: (7²)³.

Answer:

Using the power of a power rule:
(7²)³ = 7^2×3 = 7⁶ = 117,649.

Question 8:

Compare: 2⁵ and 5². Which is greater?

Answer:

Calculate both:
2⁵ = 32
5² = 25
So, 2⁵ is greater than 5².

Question 9:

What is the reciprocal of 4^-2?

Answer:

4^-2 = 1/4² = 1/16
The reciprocal of 1/16 is 16.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:

Simplify: (2³ × 3²) ÷ (2 × 3).

Answer:

First, calculate the values inside the brackets:
2³ = 8
3² = 9
Now, multiply them: 8 × 9 = 72
Denominator: 2 × 3 = 6
Now, divide: 72 ÷ 6 = 12
So, the simplified form is 12.

Question 2:

What is the value of (10⁰ + 5⁰) × 2⁰?

Answer:

Any non-zero number raised to the power of 0 is 1.
So, 10⁰ = 1
5⁰ = 1
2⁰ = 1
Now, add the first two: 1 + 1 = 2
Multiply by the third term: 2 × 1 = 2
Final answer is 2.

Question 3:

Compare 4³ and 3⁴ and state which is greater.

Answer:

First, calculate both exponents:
4³ = 4 × 4 × 4 = 64
3⁴ = 3 × 3 × 3 × 3 = 81
Now, compare the two results: 81 > 64
Therefore, 3⁴ is greater than 4³.

Question 4:

Express 0.0001 in exponential form with base 10.

Answer:

0.0001 can be written as a fraction: 1/10,000.
10,000 is 10⁴, so 1/10,000 = 10^-4.
Thus, 0.0001 in exponential form is 10^-4.

Question 5:

What is the value of (10⁰ + 5⁰) × 2²?

Answer:

Any non-zero number raised to the power of 0 is 1.
So, 10⁰ = 1 and 5⁰ = 1.
Now, add them: 1 + 1 = 2.
Next, calculate 2² = 4.
Multiply the results: 2 × 4 = 8.

Question 6:

Compare 4³ and 3⁴ using the greater than (>), less than (<), or equal to (=) sign.

Answer:

First, calculate both exponents.
4³ = 4 × 4 × 4 = 64.
3⁴ = 3 × 3 × 3 × 3 = 81.
Since 64 is less than 81, we write: 4³ < 3⁴.

Question 7:

Write the standard form of 0.0000567 using exponents.

Answer:

To write 0.0000567 in standard form using exponents:
Move the decimal point to the right until it is after the first non-zero digit (5).
This requires moving the decimal 5 places.
So, the standard form is 5.67 × 10^-5.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:

Express 256 as a power of 2. Explain each step using the laws of exponents.

Answer:

Introduction

We studied that numbers can be written as powers using exponents. Here, we express 256 as a power of 2.

Argument 1

First, we divide 256 by 2 repeatedly: 256 ÷ 2 = 128, 128 ÷ 2 = 64, 64 ÷ 2 = 32, 32 ÷ 2 = 16, 16 ÷ 2 = 8, 8 ÷ 2 = 4, 4 ÷ 2 = 2, 2 ÷ 2 = 1.

Argument 2

Since we divided 8 times, 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2⁸.

Conclusion

Thus, 256 can be expressed as 2⁸ using the laws of exponents.

Question 2:

Compare 10⁵ and 5¹⁰ using exponents. Which is greater? Show your reasoning.

Answer:

Introduction

Our textbook shows how to compare numbers written in exponential form. Here, we compare 10⁵ and 5¹⁰.

Argument 1

Calculate 10⁵ = 10 × 10 × 10 × 10 × 10 = 100,000.

Argument 2

Calculate 5¹⁰ = 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5 = 9,765,625.

Conclusion

Since 9,765,625 > 100,000, 5¹⁰ is greater than 10⁵.

Question 3:

A bacteria doubles every hour. If there are 5 bacteria initially, how many will there be after 6 hours? Use exponents to solve.

Answer:

Introduction

We studied exponential growth in real-life situations like bacteria doubling. Here, we calculate bacteria count after 6 hours.

Argument 1

Initial bacteria = 5. Since they double every hour, the growth follows 5 × 2ⁿ, where n = hours.

Argument 2

After 6 hours, bacteria count = 5 × 2⁶ = 5 × 64 = 320.

Conclusion

Thus, after 6 hours, there will be 320 bacteria.

Question 4:

Express 72 as a product of powers of prime factors. Verify your answer using the exponent laws.

Answer:

Introduction

We studied that any number can be expressed as a product of prime factors using exponents. Here, we break down 72 into its prime factors.

Argument 1

Prime factorization of 72: 72 = 2 × 2 × 2 × 3 × 3
Exponential form: 72 = 2³ × 3²

Argument 2

Verification using exponent laws: 2³ × 3² = 8 × 9 = 72, which matches the original number.

Conclusion

Thus, 72 can be written as 2³ × 3², and the calculation confirms its correctness.

Question 5:

Compare 5⁴ and 4⁵ using the properties of exponents. Which is greater and why?

Answer:

Introduction

Our textbook shows that comparing numbers in exponential form requires evaluating or simplifying them. Here, we compare 5⁴ and 4⁵.

Argument 1

Calculate 5⁴ = 5 × 5 × 5 × 5 = 625
Calculate 4⁵ = 4 × 4 × 4 × 4 × 4 = 1024

Argument 2

Since 1024 > 625, 4⁵ is greater than 5⁴. This shows that even with a smaller base, a higher exponent can result in a larger value.

Conclusion

Thus, 4⁵ > 5⁴, demonstrating the impact of exponents on magnitude.

Question 6:

A bacteria culture doubles every hour. If there are 500 bacteria initially, how many will there be after 6 hours? Use exponents to solve.

Answer:

Introduction

We studied exponential growth in real-life scenarios like population growth. Here, bacteria double every hour, forming a pattern.

Argument 1

Initial count = 500
Growth rate: doubles every hour → 2ⁿ (n = hours)

Argument 2

After 6 hours, count = 500 × 2⁶ = 500 × 64 = 32,000 bacteria.

Conclusion

Exponents help model rapid growth, showing 32,000 bacteria after 6 hours.

Question 7:

Express 343 as a power of 7. Explain the steps and verify your answer.

Answer:

Introduction

We studied that exponents represent repeated multiplication. Here, we express 343 as a power of 7.

Argument 1

We know 7 × 7 × 7 = 343.
This can be written as 7³.

Argument 2

Verification: 7³ = 7 × 7 × 7 = 343, which matches the given number.

Conclusion

Thus, 343 is correctly expressed as 7³.

Question 8:

Compare 2⁵ and 5² using exponents. Which is greater and why?

Answer:

Introduction

Our textbook shows how to compare numbers using exponents. Let’s compare 2⁵ and 5².

Argument 1

2⁵ = 2 × 2 × 2 × 2 × 2 = 32.
5² = 5 × 5 = 25.

Argument 2

Since 32 > 25, 2⁵ is greater than 5².

Conclusion

Thus, 2⁵ is greater due to higher repeated multiplication.

Question 9:

A bacteria doubles every hour. If there are 5 bacteria initially, how many will there be after 6 hours? Use exponents.

Answer:

Introduction

We studied exponential growth in real-life scenarios like bacteria doubling.

Argument 1

Initial bacteria = 5.
After 1 hour = 5 × 2 = 10.
After 6 hours = 5 × 2⁶.

Argument 2

Calculation: 2⁶ = 64, so total bacteria = 5 × 64 = 320.

Conclusion

After 6 hours, there will be 320 bacteria.

Question 10:

Express 125 as a power of 5 and explain the steps. How is this useful in simplifying calculations?

Answer:

Introduction

We studied that exponents help write large numbers compactly. Here, we express 125 as a power of 5.

Argument 1

125 can be written as 5 × 5 × 5.
Using exponents, this becomes 5³.

Argument 2

Our textbook shows how exponents simplify calculations. For example, (5³) × (5²) = 5⁵ = 3125, avoiding lengthy multiplication.

Conclusion

Exponents make math easier by reducing steps and errors.

Question 11:

Compare 2⁴ and 4². Which is greater? Justify your answer with steps.

Answer:

Introduction

We learned to compare numbers using exponents. Let’s evaluate 2⁴ and 4².

Argument 1

2⁴ = 2 × 2 × 2 × 2 = 16.
4² = 4 × 4 = 16.

Argument 2

Both values are equal. Our textbook shows similar examples like 3² and 9¹, where 9 = 9.

Conclusion

Here, 2⁴ = 4², proving exponents can represent the same value differently.

Question 12:

Express the number 1728 as a power of its prime factors and explain the process step-by-step.

Answer:

To express 1728 as a power of its prime factors, we perform prime factorization.

Step 1: Divide 1728 by the smallest prime number, which is 2.
1728 ÷ 2 = 864

Step 2: Continue dividing by 2 until it is no longer divisible.
864 ÷ 2 = 432
432 ÷ 2 = 216
216 ÷ 2 = 108
108 ÷ 2 = 54
54 ÷ 2 = 27

Step 3: Now, divide by the next smallest prime number, which is 3.
27 ÷ 3 = 9
9 ÷ 3 = 3
3 ÷ 3 = 1

Now, count the number of times each prime factor appears:

2 appears 6 times (2⁶)
3 appears 3 times (3³)

Thus, the prime factorization of 1728 is:

1728 = 2⁶ × 3³

Question 13:

Compare the values of 5³ and 3⁵ and explain which one is greater and why.

Answer:

To compare 5³ and 3⁵, we first calculate their values step-by-step.

Step 1: Calculate 5³.
5³ = 5 × 5 × 5 = 125

Step 2: Calculate 3⁵.
3⁵ = 3 × 3 × 3 × 3 × 3 = 243

Now, compare the two results:

125 (5³) < 243 (3⁵)

Reason: Even though the base (5) is larger in 5³, the exponent (5) in 3⁵ is significantly higher, leading to a much larger value. Exponents have a greater impact on the final value than the base when the exponent is sufficiently large.

Question 14:

Express the number 1728 as a power of its prime factors and explain the steps involved in the process.

Answer:

To express 1728 as a power of its prime factors, we follow these steps:

Step 1: Prime Factorization

First, we break down 1728 into its prime factors:

1728 ÷ 2 = 864
864 ÷ 2 = 432
432 ÷ 2 = 216
216 ÷ 2 = 108
108 ÷ 2 = 54
54 ÷ 2 = 27
27 ÷ 3 = 9
9 ÷ 3 = 3
3 ÷ 3 = 1

Step 2: Count the Exponents

Now, count how many times each prime number appears:

2 appears 6 times (2 × 2 × 2 × 2 × 2 × 2 = 2⁶)
3 appears 3 times (3 × 3 × 3 = 3³)

Step 3: Express in Exponential Form

Combine the prime factors with their exponents:

1728 = 2⁶ × 3³

Verification: To ensure correctness, multiply the factors back:
2⁶ = 64
3³ = 27
64 × 27 = 1728 (which matches the original number).

Question 15:

Compare the values of (5² × 5³) and (5⁵) using the laws of exponents. Explain why they are equal or different.

Answer:

To compare (5² × 5³) and (5⁵), we use the laws of exponents:

Step 1: Apply the Product of Powers Law

According to the law, when multiplying two numbers with the same base, we add their exponents:
5² × 5³ = 5^{(2 + 3)} = 5⁵

Step 2: Compare with 5⁵

Now, we see that both expressions simplify to the same value:
(5² × 5³) = 5⁵
and (5⁵) = 5⁵

Conclusion

Both expressions are equal because of the Product of Powers Law, which states that a^m × aⁿ = a^{(m + n)} when the base (a) is the same.

Verification: Calculate both sides numerically:
5² × 5³ = 25 × 125 = 3125
5⁵ = 3125
Both results are identical, confirming the law.

Question 16:

Express the number 1728 as a power of its prime factors and explain the process step-by-step.

Answer:

To express 1728 as a power of its prime factors, we perform prime factorization:

The prime factorization of 1728 is 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3, which can be written in exponential form as 2⁶ × 3³.

Value-added information: 1728 is also known as a perfect cube because it can be expressed as 12³ (12 × 12 × 12).

Question 17:

Compare the values of 5⁴ and 4⁵ using exponents and explain which is greater and why.

Answer:

To compare 5⁴ and 4⁵, we first calculate their values step-by-step:

Step 1: Calculate 5⁴.
5⁴ = 5 × 5 × 5 × 5 = 625

Step 2: Calculate 4⁵.
4⁵ = 4 × 4 × 4 × 4 × 4 = 1024

From the calculations, 4⁵ (1024) is greater than 5⁴ (625).

Reason: Even though the base (4) is smaller than 5, the higher exponent (5) in 4⁵ results in a larger value because the repeated multiplication compensates for the smaller base.

Application: This demonstrates that exponents have a significant impact on the magnitude of a number, sometimes more than the base itself.

Question 18:

Express the number 1728 as a power of its prime factors and explain the laws of exponents used in the process.

Answer:

To express 1728 as a power of its prime factors, we first perform prime factorization:

1728 ÷ 2 = 864
864 ÷ 2 = 432
432 ÷ 2 = 216
216 ÷ 2 = 108
108 ÷ 2 = 54
54 ÷ 2 = 27
27 ÷ 3 = 9
9 ÷ 3 = 3
3 ÷ 3 = 1

Thus, the prime factorization of 1728 is 2⁶ × 3³.

Here, we use the law of exponents which states that when multiplying powers with the same base, we add the exponents. However, in this case, we are breaking down the number into its prime factors.

Additionally, 1728 can also be written as 12³, since 12 × 12 × 12 = 1728. This demonstrates the power of a product rule: (a × b)ⁿ = aⁿ × bⁿ.

Question 19:

Compare and contrast the expressions 5³ × 5⁴ and (5³)⁴ using the laws of exponents. Show step-by-step simplification.

Answer:

Let's compare the two expressions using the laws of exponents:

Expression 1: 5³ × 5⁴

Using the product of powers rule: a^m × aⁿ = a^m+n
So, 5³ × 5⁴ = 5³⁺⁴ = 5⁷

Expression 2: (5³)⁴

Using the power of a power rule: (a^m)ⁿ = a^m×n
So, (5³)⁴ = 5^3×4 = 5¹²

Comparison:

5³ × 5⁴ simplifies to 5⁷, which means multiplying the bases and adding the exponents.
(5³)⁴ simplifies to 5¹², which means raising a power to another power by multiplying the exponents.

Thus, the two expressions are fundamentally different in their simplification processes and results.

Question 20:

Express 125 as a power of 5 and explain the steps involved in converting a number into its exponential form.

Answer:

To express 125 as a power of 5, we need to find the exponent to which 5 must be raised to get 125.

Step 1: Start by dividing 125 by 5 repeatedly until the quotient is 1.

Step 2: 125 ÷ 5 = 25

Step 3: 25 ÷ 5 = 5

Step 4: 5 ÷ 5 = 1

Since we divided by 5 three times, the exponent is 3.

Therefore, 125 = 5³.

Key Concept: Any number can be expressed in exponential form by identifying the base and counting how many times it is multiplied by itself.

Question 21:

Compare the values of 2⁴ and 4² and explain which one is greater. Also, generalize the rule for comparing exponents with different bases.

Answer:

To compare 2⁴ and 4², let's calculate their values step by step.

Step 1: Calculate 2⁴ = 2 × 2 × 2 × 2 = 16

Step 2: Calculate 4² = 4 × 4 = 16

Here, both values are equal, i.e., 2⁴ = 4² = 16.

General Rule: When comparing exponents with different bases, it is not always possible to determine which is greater without calculation. However, if the bases are related (like 2 and 4, where 4 = 2²), we can rewrite them with the same base for easier comparison.

For example, 4² can be written as (2²)² = 2⁴, making the comparison straightforward.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:

A bacteria culture doubles every hour. If there are 100 bacteria initially, find the number after 5 hours using exponents. Explain the steps.

Answer:

Problem Interpretation

We studied exponential growth in our textbook. Here, bacteria double every hour, so we use powers of 2.

Mathematical Modeling

Initial count: 100
Growth rate: 2ⁿ (n = hours)

Solution

After 5 hours, count = 100 × 2⁵ = 100 × 32 = 3,200 bacteria.

Question 2:

Simplify and express in exponential form: (5³ × 5⁴) ÷ 5². Show the laws used.

Answer:

Problem Interpretation

Our textbook shows how to simplify expressions using exponent laws.

Mathematical Modeling

Law 1: a^m × aⁿ = a^m+n
Law 2: a^m ÷ aⁿ = a^m-n

Solution

Step 1: 5³⁺⁴ = 5⁷
Step 2: 5^7-2 = 5⁵ (final answer).

Question 3:

A bacteria culture doubles every hour. If there are 1000 bacteria initially, how many will there be after 5 hours? Explain using exponents.

Answer:

Problem Interpretation

We studied exponential growth in our textbook. Here, bacteria double every hour, so we use exponents to model the growth.

Mathematical Modeling

Initial count = 1000. Growth rate = 2 (doubling). Time = 5 hours.

Solution

Final count = 1000 × 2⁵ = 1000 × 32 = 32,000 bacteria.

Question 4:

A notebook has 64 pages. Each page has 64 lines, and each line has 64 words. Express the total words using exponents and find the value.

Answer:

Problem Interpretation

Our textbook shows how exponents simplify repeated multiplication. Here, pages, lines, and words multiply.

Mathematical Modeling

Total words = Pages × Lines × Words = 64 × 64 × 64.

Solution

64 = 2⁶, so total words = (2⁶)³ = 2¹⁸ = 262,144 words.

Question 5:

Simplify and express in exponential form: (5² × 5³) ÷ 5⁴. Show each step.

Answer:

Problem Interpretation

Our textbook shows how to simplify expressions using laws of exponents.

Mathematical Modeling

We apply the rules: a^m × aⁿ = a^m+n and a^m ÷ aⁿ = a^m-n.

Solution

Step 1: 5² × 5³ = 5²⁺³ = 5⁵
Step 2: 5⁵ ÷ 5⁴ = 5^5-4 = 5¹ = 5

Question 6:

A bacteria culture doubles every hour. If there are 100 bacteria initially, find the number of bacteria after 5 hours using exponents. Explain the steps.

Answer:

Problem Interpretation

We studied that exponential growth occurs when a quantity doubles repeatedly. Here, bacteria count doubles every hour.

Mathematical Modeling

Initial count = 100
Growth rate = 2 (doubling)
Time = 5 hours

Solution

Final count = 100 × 2⁵ = 100 × 32 = 3,200 bacteria. Our textbook shows similar problems with population growth.

Question 7:

Simplify and express in exponential form: (5² × 5³) ÷ 5⁴. Show each step using laws of exponents.

Answer:

Problem Interpretation

We need to simplify using the laws of exponents we learned: multiplication (a^m × aⁿ = a^m+n) and division (a^m ÷ aⁿ = a^m−n).

Mathematical Modeling

Step 1: Multiply 5² × 5³ = 5²⁺³ = 5⁵
Step 2: Divide by 5⁴: 5⁵⁻⁴ = 5¹

Solution

Final answer = 5¹ (or just 5). Our textbook shows similar simplification examples.

Question 8:

A bacteria culture doubles every hour. If there are 1000 bacteria initially, find the number after 5 hours. Explain using exponents.

Answer:

Problem Interpretation

We studied exponential growth in our textbook. Here, bacteria double every hour, so we use powers of 2.

Mathematical Modeling

Initial count = 1000. Growth rate = 2ⁿ, where n = hours.

Solution

After 5 hours: 1000 × 2⁵
2⁵ = 32
Total bacteria = 1000 × 32 = 32,000

Question 9:

Simplify and express in exponential form: (5³ × 5⁷) ÷ 5⁸. Show steps.

Answer:

Problem Interpretation

Our textbook shows how to simplify expressions using laws of exponents.

Mathematical Modeling

We apply the rules: a^m × aⁿ = a^m+n and a^m ÷ aⁿ = a^m−n.

Solution

Step 1: 5³ × 5⁷ = 5¹⁰
Step 2: 5¹⁰ ÷ 5⁸ = 5²
Final answer: 5²

Question 10:

A bacteria culture doubles every hour. If there are 1000 bacteria initially, how many will there be after 5 hours? Express the answer in exponential form.

Answer:

Problem Interpretation

We studied exponential growth in our textbook. Here, bacteria double every hour, so the growth follows powers of 2.

Mathematical Modeling

Initial count: 1000
Growth rate: 2ⁿ (n = hours)

Solution

After 5 hours, the count = 1000 × 2⁵ = 1000 × 32 = 32,000. The exponential form is 10³ × 2⁵.

Question 11:

Simplify and express as a single exponent: (5³ × 5⁴) ÷ 5². Justify each step using laws of exponents.

Answer:

Problem Interpretation

Our textbook shows how to simplify expressions using exponent laws like a^m × aⁿ = a^m+n and a^m ÷ aⁿ = a^m−n.

Mathematical Modeling

Step 1: Multiply 5³ × 5⁴ = 5⁷
Step 2: Divide 5⁷ ÷ 5² = 5⁵

Solution

Final simplified form is 5⁵. We applied the laws correctly.

Question 12:

A bacteria culture doubles every hour. If there are 1000 bacteria initially, find the number of bacteria after 5 hours using exponents. Explain the steps.

Answer:

Problem Interpretation

We studied exponential growth in our textbook. Here, bacteria double every hour, so we use exponents to model the growth.

Mathematical Modeling

Initial count: 1000
Growth rate: 2 (doubling)
Time: 5 hours

Solution

Final count = 1000 × 2⁵ = 1000 × 32 = 32,000 bacteria.

Question 13:

Simplify and express in exponential form: (5² × 5³) ÷ 5⁴. Show each step as per laws of exponents.

Answer:

Problem Interpretation

Our textbook shows how to simplify expressions using exponent laws like a^m × aⁿ = a^m+n and a^m ÷ aⁿ = a^m-n.

Mathematical Modeling

Step 1: Multiply 5² × 5³ = 5²⁺³ = 5⁵
Step 2: Divide by 5⁴ = 5^5-4 = 5¹

Solution

Final simplified form is 5¹ or just 5.

Question 14:

A bacteria culture doubles every hour. If there are 1000 bacteria initially, find the number after 5 hours. Explain using exponents.

Answer:

Problem Interpretation

We studied exponential growth in bacteria. The population doubles every hour, starting with 1000.

Mathematical Modeling

Final population = Initial × 2^time. Here, time = 5 hours.

Solution

Number of bacteria = 1000 × 2⁵ = 1000 × 32 = 32,000.

Question 15:

Simplify and express in exponential form: (5² × 5³) ÷ 5⁴. Show steps using laws of exponents.

Answer:

Problem Interpretation

Our textbook shows how to simplify expressions using exponent rules.

Mathematical Modeling

First multiply (a^m × aⁿ = a^m+n), then divide (a^m ÷ aⁿ = a^m−n).

Solution

(5²⁺³) ÷ 5⁴ = 5⁵⁻⁴ = 5¹.

Question 16:

A bacteria culture doubles every hour. If there are 500 bacteria at the start, answer the following:

How many bacteria will there be after 5 hours?
Express the final number in exponential form.

Answer:

Step 1: Understand the problem.
The bacteria culture doubles every hour, which means it grows exponentially. The initial count is 500.

Step 2: Use the exponential growth formula.
Final amount = Initial amount × (2)ⁿ, where n is the number of hours.

Step 3: Plug in the values.
Final amount = 500 × (2)⁵
= 500 × 32
= 16,000 bacteria.

Step 4: Express in exponential form.
The final count is 1.6 × 10⁴ in scientific notation.

Key concept: Exponential growth describes processes where quantities multiply by a fixed factor over equal intervals.

Question 17:

A school is organizing a tree-planting drive. Each student plants 3 trees, and each tree absorbs 2³ kg of CO₂ annually. If there are 2⁴ students, calculate:

Total trees planted.
Total CO₂ absorbed annually in exponential form.

Answer:

Step 1: Calculate total trees planted.
Each student plants 3 trees, and there are 2⁴ (16) students.
Total trees = 3 × 2⁴
= 3 × 16
= 48 trees.

Step 2: Determine CO₂ absorbed per tree.
Each tree absorbs 2³ (8) kg of CO₂ annually.

Step 3: Calculate total CO₂ absorbed.
Total CO₂ = 48 × 8
= 384 kg.

Step 4: Express in exponential form.
384 kg = 3.84 × 10² kg.

Application: Exponents simplify large calculations, like environmental impact assessments.

Question 18:

A bacteria culture doubles every hour. If the initial count is 5,000 bacteria, answer the following:

Write the exponential expression for the bacteria count after n hours.
Calculate the count after 4 hours.

Answer:

The exponential expression for the bacteria count after n hours is given by: Initial Count × (2)ⁿ.
Here, the initial count is 5,000.
So, the expression becomes: 5,000 × (2)ⁿ.

To find the count after 4 hours (n = 4):
5,000 × (2)⁴
= 5,000 × 16
= 80,000 bacteria.

Note: The doubling effect demonstrates the power of exponents in real-life growth scenarios.

Question 19:

A village has a population of 10,000, which grows at a rate of 10% annually. Answer the following:

Express the population after t years using exponents.
Find the population after 3 years.

Answer:

The population after t years can be expressed using the formula: Initial Population × (1 + Rate)^t.
Here, the initial population is 10,000, and the rate is 10% or 0.10.
So, the expression becomes: 10,000 × (1.10)^t.

To find the population after 3 years (t = 3):
10,000 × (1.10)³
= 10,000 × 1.331
= 13,310 people.

Note: This shows how exponents model compound growth, such as in populations or investments.

Question 20:

A bacteria culture doubles every hour. If there are 500 bacteria at the start, calculate the number of bacteria after 5 hours using exponents. Also, write the exponential form representing the growth.

Answer:

The bacteria culture doubles every hour, which means it grows exponentially. The initial number of bacteria is 500.

After 1 hour: 500 × 2 = 1000 bacteria
After 2 hours: 500 × 2² = 2000 bacteria
After 3 hours: 500 × 2³ = 4000 bacteria
After 4 hours: 500 × 2⁴ = 8000 bacteria
After 5 hours: 500 × 2⁵ = 16000 bacteria

The exponential form representing the growth is: Number of bacteria = 500 × 2ⁿ, where n is the number of hours.

Question 21:

A village has a population of 10,000. Each person donates ₹10 to a charity fund. Express the total donation amount in exponential form and standard form.

Answer:

Each person donates ₹10, and the village has a population of 10,000.

Total donation = Number of people × Donation per person
= 10,000 × 10
= 1,00,000

In exponential form, 1,00,000 can be written as 10⁵ because:
10 × 10 × 10 × 10 × 10 = 1,00,000

In standard form, it is written as 1 × 10⁵.

Question 22:

A bacteria culture doubles every hour. If there are 500 bacteria at the start, calculate the number of bacteria after 5 hours using exponents. Also, write the final answer in exponential form.

Answer:

The problem involves exponential growth since the bacteria double every hour. The initial count is 500, and the growth rate is doubling (which means multiplying by 2 every hour).

Initial bacteria count = 500
Growth rate = 2 (doubling every hour)
Time = 5 hours

The formula for exponential growth is: Final Amount = Initial Amount × (Growth Rate)^Time

Substituting the values:
Final Amount = 500 × 2⁵
Calculate 2⁵:
2⁵ = 2 × 2 × 2 × 2 × 2 = 32

Now multiply:
Final Amount = 500 × 32 = 16,000 bacteria

Final answer in exponential form: 16,000 = 5 × 10³ × 2⁵

Question 23:

A village has a population of 10,000. Due to a health campaign, the population grows by 10% every year. Express the population after 3 years in exponential form and calculate the exact number.

Answer:

The population grows by 10% annually, which means it multiplies by 1.10 (100% + 10%) each year. This is an example of exponential growth.

Initial population = 10,000
Growth rate per year = 10% or 1.10
Time = 3 years

The formula for exponential growth is: Final Population = Initial Population × (1 + Growth Rate)^Time

Substituting the values:
Final Population = 10,000 × (1.10)³

Calculate (1.10)³:
1.10 × 1.10 = 1.21
1.21 × 1.10 = 1.331

Now multiply:
Final Population = 10,000 × 1.331 = 13,310

Final answer in exponential form: 13,310 = 10⁴ × 1.331

Question 24:

A village has a population of 10,000. Due to a health campaign, the population grows at a rate of 10% every year. Express the population after 3 years in exponential form and calculate the actual population.

Answer:

The population grows at 10% every year, so the growth factor is 1.10 (100% + 10% = 110% or 1.10).

Exponential form for population after n years: 10,000 × (1.10)ⁿ.

After 1 year: 10,000 × 1.10 = 11,000
After 2 years: 10,000 × (1.10)² = 12,100
After 3 years: 10,000 × (1.10)³ = 13,310

The actual population after 3 years is 13,310.

Exponents and Powers – CBSE NCERT Study Resources

Overview of the Chapter: Exponents and Powers

Key Concepts

Laws of Exponents

Standard Form

Applications of Exponents

Solved Examples

Practice Questions

All Question Types with Solutions – CBSE Exam Pattern

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)