Exponents and Powers

Mass of earth is 5,970,000,000,000,000,000,000,000 kg. Try reading that number aloud, counting the zeros as you go, and you will see immediately why nobody writes it that way. In Class 7 you learned the fix: exponents let us write this same mass as $5.97 \times 10^{24}$ kg, read as "5.97 times 10 raised to the power 24." That trick handled large numbers built from positive whole-number powers. This chapter asks the natural next question: what should $a^m$ mean when $m$ is not a positive whole number at all, when it is zero, or negative? The answer turns out to be beautifully consistent with everything you already know. Continuing the pattern $10^2 = 100$ , $10^1 = 10$ , $10^0 = 1$ one more step downward forces $10^{-1}$ to equal $1/10$ , and in general $a^{-m} = 1/a^m$ for any non-zero integer $a$ — a negative exponent is not a negative number, it is an instruction to take a reciprocal. Once that single idea is in place, every law of exponents you learned for positive powers — $a^m \times a^n = a^{m+n}$ , $(a^m)^n = a^{mn}$ , and the rest — turns out to hold for any integer exponents, positive, negative or zero, with no exceptions and no new rules to memorise. The real payoff arrives in the second half of the chapter. Negative exponents are exactly the tool needed to tame the other extreme of the number line: numbers too small to write comfortably, like the $0.000007$ m diameter of a red blood cell or the $0.0000000000085$ m size of a hydrogen atom. Writing such numbers in standard form, $k \times 10^n$ with $1 \le k < 10$ , turns clumsy strings of zeros into a compact, comparable format — and lets you instantly see, for instance, that the Sun's diameter is about a hundred times the Earth's, just by comparing two exponents. By the end of this chapter you will move fluently between the ordinary and the exponential form of any number, however large or small, and use the laws of exponents to simplify expressions that would otherwise take pages of multiplication.

Key takeaways

Negative exponents mean reciprocals, not negative numbers. For any non-zero integer $a$ and positive integer $m$ , $a^{-m} = 1/a^m$ ; equivalently $a^{-m}$ is the multiplicative inverse of $a^m$ .
The pattern that generates the definition. Powers of a fixed base fall by a factor of the base each time the exponent drops by 1: $10^2=100 \to 10^1=10 \to 10^0=1 \to 10^{-1}=1/10 \to 10^{-2}=1/100$ , and so on.
Any non-zero number to the power zero is 1. $a^0 = 1$ for every non-zero integer $a$ ; the base $0^0$ is left undefined.
All six laws of exponents hold for every integer exponent, not just positive ones: $a^m \times a^n = a^{m+n}$ , $a^m \div a^n = a^{m-n}$ , $(a^m)^n = a^{mn}$ , $a^m \times b^m = (ab)^m$ , $a^m/b^m = (a/b)^m$ , and $a^0=1$ .
Decimals expand with negative powers of ten. A number like $1425.36$ expands as $1 \times 10^3 + 4\times10^2 + 2\times10^1 + 5\times10^0 + 3\times10^{-1} + 6\times10^{-2}$ .
Standard form tames both extremes. Any number can be written as $k \times 10^n$ with $1 \le k < 10$ ; large numbers give positive $n$ , numbers smaller than 1 give negative $n$ .
Comparing standard-form numbers is just comparing exponents. Dividing two numbers in standard form uses the quotient law directly, so the exponent gap alone tells you the order of magnitude before you even do the decimal arithmetic.
Adding or subtracting standard-form numbers needs a common exponent first, exactly like adding fractions needs a common denominator.
Key idea: extending the pattern of powers below the exponent zero forces the definition a^-m = 1/a^m, after which every familiar law of exponents holds for all integers, and this is precisely the tool that lets any number, however large or small, be written compactly in standard form k times 10 to the power n.

The core insight

Continuing the pattern of powers of a base one exponent below zero forces the definition $a^{-m} = 1/a^m$ , which makes every law of exponents valid for all integer exponents and gives the compact standard form $k \times 10^n$ used to write and compare numbers of any size.

What this chapter covers

Section 10.1 recaps positive integer exponents as repeated multiplication, using the mass of the earth as the "too many zeros" example, and asks what a negative exponent should mean.
Section 10.2 builds negative exponents from a pattern of successive division by the base, arrives at $a^{-m} = 1/a^m$ , and uses it to expand decimals such as $1425.36$ using powers of ten, including negative ones.
Section 10.3 extends all the laws of exponents from Class 7 — product, quotient, power of a power, power of a product, power of a quotient, and the zero exponent — to any integer exponent, working through NCERT's own Examples 1 to 7.
Section 10.4 puts negative exponents to their most important use: writing very small numbers, alongside very large ones, in standard form $k \times 10^n$ , and Section 10.4.1 compares and combines numbers written this way, following NCERT's own Examples 8 and 9.

The core ideas

From repeated multiplication to negative exponents

Recall from Class 7 that for a non-zero integer $a$ and a positive integer $m$ , the power $a^m$ means $a$ multiplied by itself $m$ times: $a^m = a \times a \times \cdots \times a$ ( $m$ times). This chapter's whole extension rests on a single, very concrete observation: watch what happens to successive powers of $10$ as the exponent falls by exactly one each time.

10^2 = 100, \quad 10^1 = 10, \quad 10^0 = 1, \quad 10^{-1} = ?, \quad 10^{-2} = ? \qquad (1)

where: each number in the sequence is one-tenth of the number before it — dividing by the base $10$ is the same operation as decreasing the exponent by $1$ . Continuing that division below $10^0=1$ gives $10^{-1} = 1 \div 10 = 1/10$ , and $10^{-2} = (1/10)\div10 = 1/100$ . Nothing new has been invented — it is the identical "divide by the base" step that already carried $10^3$ down to $10^2$ , pushed one step further. The same pattern with base $3$ gives $3^{-1}=1/3$ and $3^{-2}=1/9$ .

Figure

A descending staircase of bars showing powers of ten from ten squared down to ten to the power minus two, each bar one tenth the height of the bar before it, illustrating that dividing by the base each step is what generates negative exponents

Written for a general non-zero integer base $a$ and a positive integer $m$ , the pattern is captured by a single definition, which is also the multiplicative inverse relationship $a^m \times a^{-m} = a^0 = 1$ .

a^{-m} = \frac{1}{a^m}. \qquad (2)

where: $a$ is any non-zero integer (the base can never be $0$ , since $1/0$ is undefined), and $m$ is a positive integer. Read the other way, $a^m = 1/a^{-m}$ , so $a^{-m}$ and $a^m$ are multiplicative inverses of each other — their product is always $1$ . For example, the multiplicative inverse of $2^{-4}$ is $2^4 = 16$ , and the multiplicative inverse of $7^{-2}$ is $7^2 = 49$ .

The laws of exponents, now valid for every integer

In Class 7 you proved six laws of exponents for positive whole-number exponents by directly expanding the repeated multiplication. The derivation below shows that once negative exponents are defined by $a^{-m} = 1/a^m$ , every one of those laws continues to hold with absolutely no change in form — only the restriction " $m$ and $n$ are positive integers" is lifted to " $m$ and $n$ are any integers."

\begin{aligned} a^m \times a^n &= a^{m+n} \\ a^m \div a^n &= a^{m-n} \\ (a^m)^n &= a^{mn} \\ a^m \times b^m &= (ab)^m \\ \frac{a^m}{b^m} &= \left(\frac{a}{b}\right)^m \\ a^0 &= 1 \end{aligned} \qquad (3)

where: $a$ and $b$ are non-zero integers, $m$ and $n$ are any integers whatsoever (positive, negative, or zero). These are the same six statements memorised in Class 7 — only the set of allowed exponents has widened. A useful reading habit: the product law adds exponents of the same base; the power-of-a-power law multiplies exponents; the last two laws combine two different bases raised to the same power. Mixing these up — writing $(a^m)^n = a^{m+n}$ instead of $a^{mn}$ — is one of the most common slips in this chapter, covered again under Common traps.

Expanding decimal numbers using powers of ten

A whole number like $1425$ was already written in expanded form in earlier classes as $1 \times 10^3 + 4 \times 10^2 + 2 \times 10^1 + 5 \times 10^0$ . Negative exponents extend this naturally past the decimal point, since each digit after the point is a further tenth of the one before it.

1425.36 = 1\times10^3 + 4\times10^2 + 2\times10^1 + 5\times10^0 + 3\times10^{-1} + 6\times10^{-2}. \qquad (4)

where: the digit $3$ sits in the tenths place, worth $3 \times 10^{-1} = 3/10$ , and the digit $6$ sits in the hundredths place, worth $6 \times 10^{-2} = 6/100$ — the same place-value idea used for whole numbers, continued below zero using the "divide by the base" pattern from Display (1). Expanding $1256.249$ the same way gives $1\times10^3+2\times10^2+5\times10^1+6\times10^0+2\times10^{-1}+4\times10^{-2}+9\times10^{-3}$ .

Standard form: writing very large and very small numbers compactly

The mass of the earth, $5{,}970{,}000{,}000{,}000{,}000{,}000{,}000{,}000$ kg, and the size of a plant cell, $0.00001275$ m, sit at opposite extremes of the number line, yet both are equally awkward to read or compare in ordinary decimal form. Standard form (scientific notation) fixes both problems with one convention.

N = k \times 10^n, \qquad 1 \le k < 10, \qquad n \in \mathbb{Z}. \qquad (5)

where: $k$ has exactly one non-zero digit before the decimal point, and $n$ is an integer, positive for numbers of $10$ or more, negative for numbers between $0$ and $1$ . To convert a large number, count how many places the point moves left to sit just after the first non-zero digit — that count is $n$ : $150{,}000{,}000{,}000 = 1.5 \times 10^{11}$ . To convert a small number, count how many places the point moves right to reach the first non-zero digit — that count is $-n$ : $0.000007 = 7 \times 10^{-6}$ , and $0.0016 = 1.6 \times 10^{-3}$ .

Figure

A single horizontal scale from ten to the power minus twelve on the left to ten to the power twelve on the right, with markers for the size of a plant cell near minus five, the thickness of a hair near minus two, the height of Mount Everest near four, and the distance to the sun near eleven, showing standard form placing wildly different quantities on one comparable line

Comparing and combining numbers in standard form

Once two quantities are in standard form, comparing them means comparing exponents first, leading digits second — the quotient law from Display (3) at work. The Sun's diameter is $1.4\times10^9$ m and the Earth's is $1.2756\times10^7$ m; dividing gives $\dfrac{1.4\times10^9}{1.2756\times10^7} = \dfrac{1.4}{1.2756}\times10^{9-7} \approx 1.10\times10^2$ , so the Sun is a little over a hundred times wider — NCERT rounds this to "about $100$ times." Addition and subtraction need the same exponent throughout, like a common denominator: the earth's mass ( $5.97\times10^{24}$ kg) plus the moon's ( $7.35\times10^{22}$ kg) needs $5.97\times10^{24}=597\times10^{22}$ first, giving $(597+7.35)\times10^{22}=604.35\times10^{22}$ kg.

Deriving the key results

Derivation APowers with negative exponents: deriving a to the minus m equals one over a to the m

The goal of this derivation is to show that $a^{-m} = 1/a^m$ is not an arbitrary rule to memorise but a forced consequence of two things already trusted: the pattern of dividing by the base each time the exponent falls by one, and the quotient law $a^p \div a^q = a^{p-q}$ proved in Class 7 for positive exponents. Seeing the definition emerge from familiar ground explains why a reciprocal, not a negative number, is the right meaning for a negative exponent.

Step 1. Fix a non-zero integer base $a$ . Since $a^n = a^{n-1} \times a$ by the definition of a power, dividing both sides by $a$ gives $a^{n-1} = a^n \div a$ : each step down the exponent divides the previous value by the base — Display (1) written algebraically instead of just for base $10$ .

Step 2. Apply this repeatedly starting from $a^1=a$ : $a^0 = a^1 \div a = 1$ , then $a^{-1} = a^0 \div a = 1/a$ , then $a^{-2} = a^{-1} \div a = 1/a^2$ . Every further division by $a$ both drops the exponent by one and adds another factor of $a$ to the denominator.

Step 3. Confirm this independently using the quotient law for a case where the second exponent exceeds the first. Take $a^2 \div a^5$ : by direct cancellation, $a^2 \div a^5 = \dfrac{a\times a}{a\times a\times a\times a\times a} = \dfrac{1}{a^3}$ ; by the quotient law's exponent subtraction, $a^2 \div a^5 = a^{2-5}=a^{-3}$ . Both describe the same division, so $a^{-3}=1/a^3$ .

Step 4. Generalise: for any positive integer $m$ , applying cancellation to $a^0 \div a^m$ gives $a^0 \div a^m = 1/a^m$ , while the quotient law gives $a^0 \div a^m = a^{0-m}=a^{-m}$ . Equating the two proves the general result, $a^{-m}=1/a^m$ , for every non-zero integer $a$ . Since $a^m \times a^{-m} = a^m\times(1/a^m)=1$ , $a^{-m}$ is precisely the multiplicative inverse of $a^m$ .

Go deeper · Derivation B · 1 minLaws of exponents extended to negative integers: deriving a to the m times a to the n equals a to the m plus n

Derivation BLaws of exponents extended to negative integers: deriving a to the m times a to the n equals a to the m plus n

The goal of this derivation is to prove that the product law $a^m \times a^n = a^{m+n}$ , established in Class 7 for positive exponents by repeated multiplication, continues to hold once $m$ or $n$ are negative integers, using only the definition $a^{-m}=1/a^m$ and the positive-exponent product law. Once secured for the product law, the same substitution technique proves every remaining law in Display (3) for all integers, which is why NCERT states them together as one family.

Step 1. Handle the case where both exponents are negative, using $2^{-3} \times 2^{-2}$ . Rewrite each factor as a reciprocal: $2^{-3}\times2^{-2} = \dfrac{1}{2^3}\times\dfrac{1}{2^2} = \dfrac{1}{2^3\times2^2}$ .

Step 2. Simplify the denominator with the positive-exponent product law: $2^3\times2^2=2^{3+2}=2^5$ , so $2^{-3}\times2^{-2}=1/2^5=2^{-5}$ . Since $-3+(-2)=-5$ , the law holds: $2^{-3}\times2^{-2}=2^{-3+(-2)}$ .

Step 3. Handle the mixed case, using $5^{-2}\times5^4$ . Rewrite the negative factor as a reciprocal: $5^{-2}\times5^4 = \dfrac{1}{5^2}\times5^4 = \dfrac{5^4}{5^2}$ . The quotient law gives $5^4/5^2=5^{4-2}=5^2$ , and since $-2+4=2$ , this again matches: $5^{-2}\times5^4=5^{-2+4}$ .

Step 4. Generalise: for any non-zero base $a$ and any integers $m,n$ , rewriting every negative exponent as a reciprocal reduces $a^m \times a^n$ to a ratio or product of positive-exponent powers, which the Class 7 laws simplify to a single power with exponent $m+n$ . The same substitution strategy applies unchanged to $a^m \div a^n$ , $(a^m)^n$ , and the rest, securing all six laws for every integer exponent at once.

Worked examples

Board level - evaluating powers with negative exponents (NCERT Example 1 and Try These pattern)

Find the value of (a) $2^{-3}$ , (b) $1/3^{-2}$ , and (c) the multiplicative inverse of $7^{-2}$ .

Step 1. Apply $a^{-m} = 1/a^m$ directly to part (a): $2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8}$ .

Step 2. For part (b), the negative power sits inside a reciprocal, so apply the definition first: $3^{-2}=1/3^2$ , and the reciprocal of that reciprocal returns the positive power, $\dfrac{1}{3^{-2}} = 3^2 = 9$ .

Step 3. For part (c), the multiplicative inverse of $x$ is the number that multiplies with $x$ to give $1$ ; since $7^{-2}\times7^2=7^0=1$ by the product law, the inverse of $7^{-2}$ is $7^2=49$ . Answer: $2^{-3}=1/8$ , $1/3^{-2}=9$ , and the multiplicative inverse of $7^{-2}$ is 49.

Board level - simplifying using the laws of exponents (NCERT Example 4 pattern)

Simplify and write the answer with a positive exponent: (a) $(2^5 \div 2^8)^5 \times 2^{-5}$ , and (b) $(-4)^{-3} \times 5^{-3} \times (-5)^{-3}$ .

Step 1. For part (a), simplify inside the bracket using the quotient law: $2^5 \div 2^8 = 2^{5-8} = 2^{-3}$ .

Step 2. Apply the power-of-a-power law to $(2^{-3})^5 = 2^{-15}$ , then combine with the remaining factor: $2^{-15}\times2^{-5} = 2^{-20} = 1/2^{20}$ .

Step 3. For part (b), all three factors share the exponent $-3$ , so group the bases using the power-of-a-product law "in reverse": $(-4)^{-3}\times5^{-3}\times(-5)^{-3} = \left[(-4)\times5\times(-5)\right]^{-3}$ .

Step 4. Multiply the bases: $(-4)\times5\times(-5)=100$ , so the expression is $100^{-3}=1/100^3=1/10^6$ (since $100=10^2$ ). Answer: (a) $1/2^{20}$ ; (b) $1/10^6$ .

Higher-order thinking (HOTS / olympiad-style) - solving for an unknown exponent, then reading an order of magnitude straight off two exponents (NCERT Example 5 and Section 10.4.1 combined)

(a) Find the integer $m$ so that $(-3)^{m+1} \times (-3)^5 = (-3)^7$ . (b) The diameter of the Sun is $1.4 \times 10^9$ m and the diameter of the Earth is $1.2756 \times 10^7$ m. Using only the quotient law and the two exponents 9 and 7 — before doing any decimal division — state the order of magnitude of how many times bigger the Sun is, then compute the precise ratio to three significant figures.

Step 1. Combine the left side of part (a) into a single power using the product law: $(-3)^{m+1}\times(-3)^5 = (-3)^{(m+1)+5} = (-3)^{m+6}$ . The equation becomes $(-3)^{m+6} = (-3)^7$ .

Step 2. Both sides share the identical base $-3$ , which is neither $1$ nor $-1$ (bases for which infinitely many exponents give the same value), so the exponents must be equal: $m+6=7$ , giving $m=1$ .

Step 3. For part (b), the quotient law gives the exponent of the ratio before any arithmetic on the leading digits: $\dfrac{1.4\times10^9}{1.2756\times10^7} = \dfrac{1.4}{1.2756}\times10^{9-7} = \dfrac{1.4}{1.2756}\times10^2$ . From the exponent gap $9-7=2$ alone, with both leading digits close to $1$ , the ratio must be of order $10^2$ — hundreds, not tens or thousands.

Step 4. Compute the leading-digit factor precisely: $1.4 \div 1.2756 \approx 1.098$ , so the ratio is $1.098\times10^2\approx109.8$ . Answer: $m=1$ ; the exponent gap alone shows the Sun's diameter is of order $10^2$ times the Earth's, and the precise ratio is about $110$ , consistent with NCERT's own rounded "about 100 times." Timing note: this pair should take about ninety seconds once both skills are automatic. The exponent-matching in part (a) only works because the base $-3$ is not $1$ , $-1$ , or $0$ — flag that condition before equating exponents. In part (b), read off the order of magnitude from the bare exponent subtraction first; diving straight into long division loses time a quick estimate would have saved for free.

Common traps

The single most frequent slip is reading $a^{-m}$ as "negative $a^m$ " — writing $2^{-3}=-8$ instead of the correct reciprocal $1/8$ ; a negative exponent never makes the value negative unless the base itself is negative with an odd exponent. A closely related error is applying the reciprocal to the wrong part of an expression, simplifying $\dfrac{1}{3^{-2}}$ as $\dfrac{1}{9}$ instead of correctly recognising it as $3^2=9$ . Confusing the product law with the power-of-a-power law is equally common: $a^m\times a^n=a^{m+n}$ adds exponents of the same base, while $(a^m)^n=a^{mn}$ multiplies them. Students also forget the product and quotient laws need the same base on both factors — $2^3\times3^2$ cannot be combined into a single power, since the bases differ. When equating exponents from $a^p=a^q$ , the base must be checked first: this fails for $a=1$ (since $1^n=1$ for every $n$ ) and for $a=-1$ with even exponents. Finally, in standard form the common slip is moving the decimal point the wrong direction — a number smaller than $1$ always needs a negative exponent, and adding or subtracting standard-form numbers is only valid once both share a common exponent.

Common trapThe single biggest misconception in this chapter: a negative exponent does not make a number negative

Every year, the most-lost marks in this chapter come from treating $a^{-m}$ as though the minus sign belonged to the value rather than to an instruction about the exponent. It does not: $2^{-3}$ is a small positive fraction, $1/8$ , not $-8$ . The correct mental model is that a negative exponent commands you to divide $1$ by the corresponding positive power of the base: $a^{-m}=1/a^m$ — exactly why $a^{-m}$ is called the multiplicative inverse of $a^m$ , their product always $1$ . The only way a power genuinely comes out negative is when the base itself is negative and the exponent is odd, as in $(-2)^{-3}=1/(-2)^3=-1/8$ — the negativity comes entirely from the base, never from the exponent alone. Whenever you see a negative exponent, your first move is always "take the reciprocal of the positive power," never "make the number negative."

You should be able to…

State $a^{-m} = 1/a^m$ for a non-zero integer $a$ and positive integer $m$ , and explain why it follows the "divide by the base" pattern below the exponent zero.
Explain why $a^{-m}$ is the multiplicative inverse of $a^m$ , and find the multiplicative inverse of any given power.
Evaluate expressions like $2^{-3}$ , $1/3^{-2}$ , and $(-5)^{-2}$ correctly, without turning a negative exponent into a negative value.
State all six laws of exponents from Display (3) and know they apply for any integer exponent, not only positive ones.
Simplify expressions combining several laws of exponents at once, expressing the final answer with a positive exponent.
Expand a decimal such as $1425.36$ using positive and negative powers of ten, matching each digit to its place value.
Convert any number into standard form $k \times 10^n$ with $1 \le k < 10$ , choosing the correct sign of $n$ , and convert it back.
Compare two standard-form numbers by subtracting exponents (quotient law) before doing any decimal arithmetic.
Add or subtract two standard-form numbers by first rewriting them with a common exponent.

Ask the tutor: if a sign or an exponent rule trips you up mid-problem, ask the tutor to rewrite just that one step using the reciprocal definition and walk through it with you.

Why is any non-zero number raised to the power zero equal to 1, and not 0?

It follows directly from the quotient law: $a^m \div a^m = a^{m-m} = a^0$ . But dividing any non-zero number by itself always gives exactly $1$ , so $a^0$ must equal $1$ too — there is no freedom to choose otherwise without breaking the quotient law. This is also visible in the descending staircase pattern: each step down from $10^2=100$ to $10^1=10$ divides by the base, and the next division, from $10^1$ down to $10^0$ , gives exactly $10/10=1$ , not $0$ . The one exception is $0^0$ , left undefined because it would involve dividing zero by zero.

Why can the base in a to the power minus m never be zero?

Because $a^{-m}$ is defined as $1/a^m$ , and if $a$ were $0$ then $a^m$ would also be $0$ for any positive $m$ , making $a^{-m}=1/0$ , which is undefined. Division by zero is never allowed, and negative exponents are built entirely out of division, so " $a$ is non-zero" is not a minor technicality — it is the reason the definition works at all. This is why every law and example in this chapter states "for any non-zero integer $a$ " rather than simply "for any integer $a$ ."

How do I decide whether the exponent in standard form should be positive or negative?

Look at whether the original number is $10$ or larger, or smaller than $1$ . If it is $10$ or more, move the decimal point left to leave one non-zero digit in front, and the count of places moved becomes a positive exponent, since the true value is larger than the digit alone suggests — $4{,}050{,}000 = 4.05\times10^6$ . If it is smaller than $1$ , move the point right to reach the first non-zero digit, and that count becomes a negative exponent — $0.000035 = 3.5\times10^{-5}$ . A quick check afterward: multiplying your answer back out should return the original number.

Do the laws of exponents from this chapter still work if the exponent is a fraction, like a half?

Not within this chapter — the definition $a^{-m}=1/a^m$ and all six laws in Display (3) are stated and proved only for integer exponents, positive, negative, or zero. Extending exponents to fractions, so that $a^{1/2}$ means the square root of $a$ , is a genuinely new idea needing its own careful definition, and is not covered by anything derived here; you will meet it properly in later classes. Until then, treat any expression with a fractional or decimal exponent as outside the scope of these rules, and do not apply $a^m\times a^n=a^{m+n}$ or the other laws to it without that extra groundwork.