Exponents And Powers — for Class 7-8

What Exponents and Powers Actually Mean

When we multiply the same number repeatedly, writing it out gets tedious fast. Instead of writing $2 \times 2 \times 2 \times 2 \times 2$ , we write $2^5$ . That’s it — that’s the whole idea.

The number being multiplied is called the base, and the count of how many times we multiply it is the exponent (also called the power or index). So in $2^5$ , the base is $2$ and the exponent is $5$ .

This notation shows up everywhere — from scientific notation in Class 8 to logarithms in Class 11 and beyond. Getting comfortable with the laws here means you’re building the foundation for every algebra chapter ahead. NCERT Class 7 Chapter 13 and Class 8 Chapter 12 together cover this fully, and questions from here appear regularly in school olympiads too.

Key Terms and Definitions

Base: The number that is being multiplied repeatedly. In $a^n$ , $a$ is the base.

Exponent / Power / Index: How many times the base is multiplied by itself. In $a^n$ , $n$ is the exponent.

Exponential Form: Writing a number using a base and exponent. Example: $81 = 3^4$ .

Expanded Form: Writing out all the multiplications. Example: $3^4 = 3 \times 3 \times 3 \times 3$ .

Perfect Power: A number that can be written as an integer base raised to an integer exponent. $64 = 2^6 = 4^3 = 8^2$ — it’s a perfect power in three different ways.

Quick Reference

Expression	Base	Exponent	Value
$5^3$	5	3	125
$(-2)^4$	-2	4	16
$(-3)^3$	-3	3	-27
$\left(\frac{2}{3}\right)^2$	$\frac{2}{3}$	2	$\frac{4}{9}$

Students often confuse $(-2)^4$ and $-2^4$ . These are different things. $(-2)^4 = (-2)\times(-2)\times(-2)\times(-2) = +16$ . But $-2^4$ means $-(2^4) = -16$ . The bracket is doing real work here.

Laws of Exponents

These seven laws are the real scoring material. Learn them cold — CBSE asks direct application questions, and olympiad problems just combine them cleverly.

Let $a$ , $b$ be any rational numbers and $m$ , $n$ be integers.

Law 1 — Product of same base:

a^m \times a^n = a^{m+n}

Law 2 — Quotient of same base:

a^m \div a^n = a^{m-n} \quad (a \neq 0)

Law 3 — Power of a power:

(a^m)^n = a^{mn}

Law 4 — Power of a product:

(ab)^n = a^n \cdot b^n

Law 5 — Power of a quotient:

\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} \quad (b \neq 0)

Law 6 — Zero exponent:

a^0 = 1 \quad (a \neq 0)

Law 7 — Negative exponent:

a^{-n} = \frac{1}{a^n} \quad (a \neq 0)

Why These Laws Work

Law 1 makes sense if you expand: $a^3 \times a^2 = (a \times a \times a) \times (a \times a) = a^5$ . You’re just counting the total number of $a$ ‘s.

Law 3 is the one students always second-guess. $(a^m)^n$ means “take $a^m$ and multiply it by itself $n$ times.” Each copy of $a^m$ contributes $m$ factors of $a$ , and there are $n$ copies, so total factors = $mn$ .

Law 6 follows directly from Law 2: $a^m \div a^m = a^{m-m} = a^0$ . But anything divided by itself equals 1. So $a^0 = 1$ .

Law 7 similarly: $a^{-n} = a^{0-n} = a^0 \div a^n = 1 \div a^n = \frac{1}{a^n}$ .

Solved Examples

Easy — CBSE Class 7

Example 1: Simplify $3^4 \times 3^2$ .

Using Law 1: $3^4 \times 3^2 = 3^{4+2} = 3^6 = 729$ .

Example 2: Write $\frac{5^7}{5^3}$ in simplest exponential form.

Using Law 2: $\frac{5^7}{5^3} = 5^{7-3} = 5^4$ .

Example 3: Find the value of $(2^3)^2$ .

Using Law 3: $(2^3)^2 = 2^{3 \times 2} = 2^6 = 64$ .

Medium — CBSE Class 8

Example 4: Simplify $\frac{2^5 \times 3^3 \times 4}{3 \times 32}$ .

Rewrite everything as prime powers: $4 = 2^2$ , $32 = 2^5$ , $3 = 3^1$ .

\frac{2^5 \times 3^3 \times 2^2}{3^1 \times 2^5} = \frac{2^{5+2} \times 3^3}{3^1 \times 2^5} = \frac{2^7 \times 3^3}{2^5 \times 3^1} = 2^{7-5} \times 3^{3-1} = 2^2 \times 3^2 = 4 \times 9 = 36

Example 5: Simplify $\left(\frac{2}{3}\right)^{-2}$ .

Using Law 7: $\left(\frac{2}{3}\right)^{-2} = \frac{1}{\left(\frac{2}{3}\right)^2} = \frac{1}{\frac{4}{9}} = \frac{9}{4}$ .

Or use the shortcut: $\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$ . So $\left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$ .

The shortcut for negative exponent with fractions — flip the fraction, make the exponent positive — saves 30 seconds in exams. Worth memorising.

Example 6: Express $0.00004$ in standard form (scientific notation).

Count how many places the decimal moves: $0.00004 = 4 \times 10^{-5}$ .

The decimal moves 5 places to the right, so we use $10^{-5}$ .

Hard — Class 8 / Olympiad

Example 7: If $2^x = 3^y = 6^{-z}$ , prove that $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$ .

Let $2^x = 3^y = 6^{-z} = k$ .

Then: $2 = k^{1/x}$ , $3 = k^{1/y}$ , $6 = k^{-1/z}$ .

Since $6 = 2 \times 3$ :

k^{-1/z} = k^{1/x} \times k^{1/y} = k^{1/x + 1/y}

So $-\frac{1}{z} = \frac{1}{x} + \frac{1}{y}$ , which gives $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$ . $\blacksquare$

Example 8: Find $x$ if $\left(\frac{3}{4}\right)^{3} \times \left(\frac{3}{4}\right)^{-6} = \left(\frac{3}{4}\right)^{2x-1}$ .

Left side: $\left(\frac{3}{4}\right)^{3+(-6)} = \left(\frac{3}{4}\right)^{-3}$ .

So: $\left(\frac{3}{4}\right)^{-3} = \left(\frac{3}{4}\right)^{2x-1}$ .

Since bases are equal: $2x - 1 = -3$ , so $2x = -2$ , giving $x = -1$ .

Exam-Specific Tips

CBSE Class 7 & 8: Questions are typically 1-mark (write in exponential form) or 2-mark (simplify using laws). The 3-mark questions usually involve two or three laws combined. Full marks require showing each step — don’t skip the intermediate line. Standard form / scientific notation is a guaranteed 1-2 marks in Class 8 boards.

NTSE & Math Olympiads: Olympiad problems combine exponent laws with number theory. Classic pattern: given $a^x = b^y = c^z$ and $abc = 1$ , find the value of $xy + yz + zx$ . The technique in Example 7 above handles the entire family of these problems.

Standard Form (Scientific Notation) — Class 8 Special Topic

A number in standard form is written as $a \times 10^n$ where $1 \leq a < 10$ and $n$ is an integer.

$47,000 = 4.7 \times 10^4$ (large number → positive exponent)
$0.0047 = 4.7 \times 10^{-3}$ (small number → negative exponent)

CBSE consistently asks students to convert between standard form and ordinary notation, and to compare two numbers written in standard form. If the powers of 10 match, compare the $a$ values. If they don’t, the larger power wins.

Common Mistakes to Avoid

Mistake 1 — Adding exponents when multiplying different bases. $2^3 \times 3^3 \neq 6^6$ . Law 1 only works when bases are the same. For $2^3 \times 3^3$ , use Law 4 in reverse: $(2 \times 3)^3 = 6^3$ .

Mistake 2 — Mishandling negative bases. $(-2)^3 = -8$ (negative, because odd power). $(-2)^4 = +16$ (positive, because even power). Odd exponent preserves the negative sign; even exponent kills it.

Mistake 3 — Thinking $a^0 = 0$ . This is wrong. $a^0 = 1$ for any non-zero $a$ . The only exception is $0^0$ , which is undefined (and won’t appear in your exam). Many students lose 1 mark on this every year.

Mistake 4 — Applying Law 3 incorrectly. $(a^m)^n = a^{mn}$ , NOT $a^{m+n}$ . A common error: $(2^3)^4 = 2^7$ ? No. It’s $2^{12}$ .

Mistake 5 — Forgetting to convert bases before applying laws. When simplifying $\frac{8^3}{4^5}$ , don’t stop here — express both in terms of 2: $\frac{(2^3)^3}{(2^2)^5} = \frac{2^9}{2^{10}} = 2^{-1} = \frac{1}{2}$ . Questions that look unsimplifiable almost always have a hidden common base.

Practice Questions

Q1. Simplify: $5^3 \times 5^{-2} \times 5^0$

$5^3 \times 5^{-2} \times 5^0 = 5^{3 + (-2) + 0} = 5^1 = 5$

Q2. Express $729$ as a power of $3$ .

$729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6$

Q3. Simplify: $\left(\frac{4}{5}\right)^{-3}$

$\left(\frac{4}{5}\right)^{-3} = \left(\frac{5}{4}\right)^3 = \frac{125}{64}$

Q4. Write $0.000000056$ in standard form.

Move the decimal 8 places to the right to get $5.6$ .

0.000000056 = 5.6 \times 10^{-8}

Q5. Simplify: $\frac{(2^3)^2 \times 3^4}{6^2 \times 4^2}$

Rewrite all in prime factors: $6^2 = (2 \times 3)^2 = 2^2 \times 3^2$ , and $4^2 = (2^2)^2 = 2^4$ .

\frac{2^6 \times 3^4}{2^2 \times 3^2 \times 2^4} = \frac{2^6 \times 3^4}{2^6 \times 3^2} = 3^{4-2} = 3^2 = 9

Q6. Find $x$ : $7^{x+1} \times 7^3 = 7^{10}$

$7^{(x+1)+3} = 7^{10}$

$x + 4 = 10$

$x = 6$

Q7. If $a = 2$ and $b = 3$ , find the value of $(a^b)^a - (b^a)^b$ .

$(a^b)^a = (2^3)^2 = 2^6 = 64$

$(b^a)^b = (3^2)^3 = 3^6 = 729$

$(a^b)^a - (b^a)^b = 64 - 729 = -665$

Q8. Arrange in ascending order: $2^{50}$ , $3^{30}$ , $5^{20}$ .

Make the exponents comparable. Express each as a power to the 10th:

$2^{50} = (2^5)^{10} = 32^{10}$

$3^{30} = (3^3)^{10} = 27^{10}$

$5^{20} = (5^2)^{10} = 25^{10}$

Now comparing $25^{10}$ , $27^{10}$ , $32^{10}$ : since $25 < 27 < 32$ , the ascending order is $5^{20} < 3^{30} < 2^{50}$ .

Q9. Simplify: $\frac{10^5 \times 10^3}{10^4 \times 10^2}$

\frac{10^{5+3}}{10^{4+2}} = \frac{10^8}{10^6} = 10^{8-6} = 10^2 = 100

Q10. The mass of the Earth is approximately $5,970,000,000,000,000,000,000,000$ kg. Express this in standard form.

Count the digits after the first significant digit: there are 24 zeros after accounting for the decimal shift.

$5,970,000,000,000,000,000,000,000 = 5.97 \times 10^{24}$ kg

Frequently Asked Questions

What is the difference between “power” and “exponent”?

They mean the same thing in school mathematics. In $3^5$ , we say “3 raised to the power 5” and the exponent is 5. Some books use “index” (plural: indices) instead. All three words — power, exponent, index — refer to the raised number.

Why is any number to the power 0 equal to 1?

The cleanest reason: consider $\frac{a^n}{a^n}$ . This equals 1 (anything divided by itself). By Law 2, it also equals $a^{n-n} = a^0$ . So $a^0 = 1$ . This works for any non-zero $a$ .

Can the exponent be a fraction or decimal?

In Class 7-8, exponents are always integers (whole numbers including negatives). Fractional exponents like $a^{1/2} = \sqrt{a}$ appear in Class 9-10. For now, focus on integer exponents.

How do I compare two numbers with different bases and different exponents?

Find a common exponent or common base if possible. If not, use logarithms (Class 11 topic) or the technique in Q8 above — convert both to a common power of 10.

What is $0^0$ ?

It’s undefined. This question sometimes appears as a trick. The reason: $0^n = 0$ for all positive $n$ suggests $0^0$ should be 0, but $a^0 = 1$ for all non-zero $a$ suggests it should be 1. The conflict means mathematicians leave it undefined. You won’t be asked to evaluate this in CBSE exams.

Is $(-1)^{100}$ positive or negative?

Positive. $(-1)$ raised to any even power is $+1$ . Raised to any odd power is $-1$ . So $(-1)^{100} = 1$ .

What is scientific notation used for?

Anywhere numbers are extremely large or extremely small — distance to stars, mass of atoms, size of bacteria, national budgets. It makes comparison and calculation much easier. Physics and Chemistry in Class 9-10 use it constantly.

How many laws do I actually need to memorise for CBSE?

All seven. In practice, Laws 1, 2, 3, and 7 appear in 80% of questions. Law 6 (zero exponent) catches students who haven’t memorised it. Laws 4 and 5 (power of a product/quotient) come up in the “express in simplest form” type questions where you need to split or combine bases.