NCERT Solutions for Class 8 Maths Chapter 10: Exponents and Powers — Free PDF

In Class 7, you learnt about exponents with positive integer powers. Chapter 10 of Class 8 extends this to negative exponents and the standard form (scientific notation) of numbers. You will also learn to apply all laws of exponents to rational number bases.

Exponents are essential for expressing very large numbers (like the distance between stars) and very small numbers (like the size of atoms) in a compact form. Scientists, engineers, and mathematicians use exponential notation every day to handle quantities that would otherwise require dozens of zeros.

This chapter has two exercises. Exercise 10.1 focuses on negative exponents, laws of exponents applied to integer and rational bases, and simplification problems. Exercise 10.2 deals with expressing numbers in standard form and converting between standard and usual form. Mastering this chapter provides a strong foundation for logarithms, scientific notation, and algebraic manipulation in higher classes.

Before diving into the exercises, let us review the fundamental ideas of this chapter.

Exponent (Power): In $a^n$, the number $a$ is the base and $n$ is the exponent (or power). It tells you how many times the base is multiplied by itself. For example, $2^5 = 2 \times 2 \times 2 \times 2 \times 2 = 32$.

Negative Exponent: A negative exponent indicates the reciprocal. $a^{-n} = \dfrac{1}{a^n}$. This is the single most important new concept in this chapter. For instance, $3^{-2} = \dfrac{1}{3^2} = \dfrac{1}{9}$.

Zero Exponent: For any non-zero number $a$, $a^0 = 1$. This follows from the quotient rule: $a^n \div a^n = a^{n-n} = a^0 = 1$.

Standard Form (Scientific Notation): A number written as $k \times 10^n$ where $1 \leq k < 10$ and $n$ is an integer. For example, the speed of light $\approx 3 \times 10^8$ m/s, and the mass of a hydrogen atom $\approx 1.67 \times 10^{-27}$ kg.

Rational Number Base: Exponent laws apply to rational numbers just as they do to integers. For example, $\left(\dfrac{2}{3}\right)^{-2} = \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}$.

For non-zero $a$ and integers $m, n$:

1. Product Rule: $a^m \times a^n = a^{m+n}$

When multiplying powers with the same base, add the exponents. Example: $5^3 \times 5^4 = 5^{3+4} = 5^7$.

2. Quotient Rule: $\dfrac{a^m}{a^n} = a^{m-n}$

When dividing powers with the same base, subtract the exponents. Example: $\dfrac{7^5}{7^2} = 7^{5-2} = 7^3$.

3. Power of a Power: $(a^m)^n = a^{mn}$

When raising a power to another power, multiply the exponents. Example: $(2^3)^4 = 2^{12}$.

4. Power of a Product: $(ab)^m = a^m \cdot b^m$

The power distributes over multiplication. Example: $(3 \times 5)^2 = 3^2 \times 5^2 = 9 \times 25 = 225$.

5. Power of a Quotient: $\left(\dfrac{a}{b}\right)^m = \dfrac{a^m}{b^m}$

The power distributes over division. Example: $\left(\dfrac{2}{7}\right)^3 = \dfrac{8}{343}$.

6. Zero Exponent: $a^0 = 1$

7. Negative Exponent: $a^{-m} = \dfrac{1}{a^m}$

These seven laws form the complete toolkit for simplifying any exponent expression. Every problem in this chapter can be solved by applying one or more of these rules.

**Q1. Evaluate: $2^{-3}$.**

Solution:

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**Q2. Simplify: $\dfrac{5^4 \times 5^{-2}}{5^3}$.**

Solution:

Step 1: Apply the product rule to the numerator.

Step 2: Apply the quotient rule.

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**Q3. Find $x$ if $\left(\dfrac{2}{3}\right)^x \times \left(\dfrac{2}{3}\right)^5 = \left(\dfrac{2}{3}\right)^{-3}$.**

Solution:

Using the product rule on the left side:

Since the bases are equal, the exponents must be equal:

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**Q4. Express $0.00000347$ in standard form.**

Solution:

Move the decimal point $6$ places to the right to get $3.47$.

The exponent is $-6$ because we moved the decimal to the right (making the number larger), so we compensate with a negative power of $10$.

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**Q5. Simplify: $\left(\dfrac{3}{4}\right)^{-3} \times \left(\dfrac{3}{4}\right)^{7}$.**

Solution:

Using the product rule:

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**Q6. Find the value of $\left(2^{-1} + 3^{-1}\right)^{-1}$.**

Solution:

Step 1: Evaluate the expression inside the bracket.

Step 2: Apply the negative exponent.

Important note: $\left(2^{-1} + 3^{-1}\right)^{-1} \neq 2 + 3$. A very common mistake is to distribute the outer exponent incorrectly.

**Q1. Simplify and express with positive exponents: $\left(\dfrac{3}{4}\right)^{-2}$.**

Solution:

A negative exponent on a fraction flips the fraction:

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**Q2. Express $4.5 \times 10^4$ as a usual number.**

Solution:

Move the decimal $4$ places to the right:

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**Q3. Simplify: $\dfrac{2^5 \times 3^2 \times 7}{2^3 \times 3^4}$.**

Solution:

Group like bases and apply the quotient rule:

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**Q4. Express $3,84,00,000$ in standard form.**

Solution:

$3,84,00,000 = 3.84 \times 10^7$

We move the decimal $7$ places to the left, so the exponent is $+7$.

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**Q5. Simplify: $\dfrac{(3^2)^3 \times 3^{-5}}{3^4 \times 3^{-7}}$.**

Solution:

Step 1: Simplify the numerator using the power-of-a-power rule.

Step 2: Simplify the denominator.

Step 3: Divide.

Here are more worked examples that cover common exam patterns.

**Example 1. Simplify: $\left(\dfrac{-2}{5}\right)^{-3} \times \left(\dfrac{-2}{5}\right)^{6}$.**

Solution:

Using the product rule:

Note: Since the exponent $3$ is odd, the negative sign is preserved.

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**Example 2. By what number should $(-3)^{-3}$ be multiplied to get $(-3)^{-1}$?**

Solution:

Let the required number be $x$.

So the required number is $9$.

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**Example 3. Express $0.00000000085$ in standard form.**

Solution:

Move the decimal $10$ places to the right to get $8.5$.

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**Example 4. Simplify: $\dfrac{(-2)^5 \times (-2)^{-3}}{(-2)^{-4} \times (-2)^2}$.**

Solution:

Numerator: $(-2)^{5+(-3)} = (-2)^2 = 4$.

Denominator: $(-2)^{-4+2} = (-2)^{-2} = \dfrac{1}{4}$.

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**Example 5. If $5^x \times 5^3 = 5^7$, find $x$. If $8^x \div 8^2 = 8^5$, find $x$.**

Solution:

For the first: $5^{x+3} = 5^7 \Rightarrow x + 3 = 7 \Rightarrow x = 4$.

For the second: $8^{x-2} = 8^5 \Rightarrow x - 2 = 5 \Rightarrow x = 7$.

Students frequently lose marks on exponent problems due to these common errors:

Mistake 1: Confusing the power of a sum with the sum of powers.
$(a + b)^2 \neq a^2 + b^2$. For example, $(2 + 3)^2 = 25$, but $2^2 + 3^2 = 4 + 9 = 13$. The correct expansion is $(a+b)^2 = a^2 + 2ab + b^2$.

Mistake 2: Distributing a negative exponent incorrectly over addition.
$(2^{-1} + 3^{-1})^{-1} \neq 2 + 3$. You must first compute the sum inside the bracket, then apply the exponent. The correct answer is $\dfrac{6}{5}$ (shown in Exercise 10.1, Q6 above).

Mistake 3: Forgetting to flip the fraction with a negative exponent.
$\left(\dfrac{2}{3}\right)^{-2} \neq \dfrac{2^{-2}}{3^{-2}}$ ... well, technically that equals $\dfrac{3^2}{2^2} = \dfrac{9}{4}$, which is correct. But students sometimes write $\dfrac{4}{9}$ by mistake. Remember: a negative exponent on a fraction flips the fraction, then applies the positive exponent.

Mistake 4: Sign errors with negative bases.
$(-3)^2 = 9$ (positive), but $(-3)^3 = -27$ (negative). Even powers make the result positive; odd powers keep it negative.

Mistake 5: Errors in standard form.
Students sometimes write $34.7 \times 10^5$ instead of $3.47 \times 10^6$. Remember, in standard form, $k$ must satisfy $1 \leq k < 10$.

Test your understanding with these additional questions.

Q1. Simplify: $\left(\dfrac{5}{7}\right)^{-2} \times \left(\dfrac{5}{7}\right)^{-5}$.

Answer: $\left(\dfrac{5}{7}\right)^{-7} = \left(\dfrac{7}{5}\right)^{7} = \dfrac{7^7}{5^7} = \dfrac{823543}{78125}$.

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Q2. Express in standard form: (a) $67,30,000$ (b) $0.0000502$.

Answer: (a) $6.73 \times 10^6$. (b) $5.02 \times 10^{-5}$.

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Q3. Find $m$: $\left(\dfrac{4}{9}\right)^3 \times \left(\dfrac{4}{9}\right)^{-6} = \left(\dfrac{4}{9}\right)^{2m-1}$.

Answer: LHS $= \left(\dfrac{4}{9}\right)^{3-6} = \left(\dfrac{4}{9}\right)^{-3}$. So $2m - 1 = -3$, giving $m = -1$.

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Q4. Simplify and write with positive exponents: $\dfrac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}$.

Answer: Write $10 = 2 \times 5$, $125 = 5^3$, $6 = 2 \times 3$.

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Q5. The distance from the Earth to the Sun is approximately $1.496 \times 10^{11}$ m. Express this as a usual number.

Answer: $1.496 \times 10^{11} = 149,600,000,000$ m (approximately $149.6$ billion metres).

• $a^{-m} = \dfrac{1}{a^m}$ — negative exponents mean reciprocals.
  - All laws of exponents (product, quotient, power rules) work for integer exponents.
  - Standard form writes numbers as $k \times 10^n$ where $1 \leq k < 10$.
  - Exponents of rational numbers follow the same rules: $\left(\dfrac{a}{b}\right)^{-n} = \left(\dfrac{b}{a}\right)^n$.
  - Always simplify step by step — convert to the same base, apply one law at a time, and simplify at the end.
  - When expressing numbers in standard form, count decimal moves carefully and check that $k$ is between $1$ and $10$.

Practise exponents on SparkEd to build fluency with these essential rules!