Pure 2.1 - Laws of indices for rational exponents | Edexcel A-Level Mathematics

Indices let you work exactly with roots, reciprocals and powers. You will explain what rational indices mean, choose the right index law and simplify numerical and algebraic expressions without losing their conditions.

Meaning of indices

How can a negative power of a tiny fraction produce a large positive number? Consider $\left(\frac1{16}\right)^{-3/4}$ . There are two instructions here: take a root and take a reciprocal. By the end, you will be able to read both at a glance.

You already know

$a^4=a\times a\times a\times a$ . Brackets identify the whole base: $(2x-1)^4$ raises all of $2x-1$ to the fourth power.

In $5^{2/3}$ , the base is $5$ and the index is $\frac23$ . In $p^{1/2}q^3$ , there are two bases; each has its own index. The index is an instruction attached to that base, not a number multiplying it.

Rational exponent

A rational exponent is an index expressible as $\frac mn$ , with integers $m$ and $n$ , where $n\ne0$ . For example, $-\frac23$ is rational; $\sqrt2$ is not.

For general algebra with rational exponents, we use positive bases so all the real roots and laws below work together. Zero can be raised to positive rational powers, but a zero or negative index at base zero is undefined. Odd roots of negative numbers can exist, such as $\sqrt[3]{-8}=-2$ ; do not extend every positive-base law to negative bases without checking its domain.

Key idea

Identify the whole base before interpreting its index; a negative index changes the reciprocal, not the sign of the value.

A useful boundary check is $-3^2=-9$ , whereas $(-3)^2=9$ : the brackets change the base.

The three core laws

Multiplying $a^3$ by $a^2$ joins three factors and two factors, giving $a^5$ . Dividing cancels factors, while $(a^3)^2$ repeats three factors twice. These observations explain which operation belongs to each law.

Core index laws

For $a>0$ and rational $m$ , $n$ ,

a^ma^n=a^{m+n},\qquad \frac{a^m}{a^n}=a^{m-n},\qquad (a^m)^n=a^{mn}

Not in the formula booklet: learn it.

The zero index follows from division: $a^m/a^m=1=a^{m-m}=a^0$ . The same rules work with fractional indices; only the fraction arithmetic changes.

Worked example 1: product, quotient and power

Question. Simplify $x^{5/3}x^{1/3}$ , $x^{7/2}/x^{3/2}$ and $(x^{-2/5})^{10}$ , where $x>0$ .

Plan. Add, subtract or multiply indices according to the operation.

\begin{aligned} x^{5/3}x^{1/3}&=x^{6/3}=x^2 &&\text{product: add}\\ x^{7/2}/x^{3/2}&=x^{4/2}=x^2 &&\text{quotient: subtract}\\ (x^{-2/5})^{10}&=x^{-4}=\frac1{x^4} &&\text{power: multiply} \end{aligned}

Why this step? A power of a power repeats the exponent; it does not join two separate powers.

Answer. $x^2$ , $x^2$ and $1/x^4$ .

Check. For the product, $x=8$ gives $32\times2=64=8^2$ . For the quotient, $x=4$ gives $128/8=16=4^2$ . For the power, $x=32$ gives $(1/4)^{10}=1/1048576=1/32^4$ .

Now change only the outer power in the third expression. A fractional outer power still multiplies the inner index.

Worked example 2: a rational outer power

Question. Simplify $(x^{-2/5})^{5/2}$ , where $x>0$ , using positive indices.

Plan. Multiply the indices, then interpret the negative index.

\begin{aligned} (x^{-2/5})^{5/2}&=x^{(-2/5)(5/2)} &&\text{power of a power}\\ &=x^{-1}=\frac1x &&\text{reciprocal} \end{aligned}

Why this step? The negative sign remains in the product of the indices; the value remains positive.

Answer. $1/x$ .

Check. At $x=32$ , $x^{-2/5}=1/4$ and $(1/4)^{5/2}=1/32$ .

A law is useful only when its structure matches the expression in front of you.

Watch out

Wrong: $x^2x^3=x^6$ . These are two powers being multiplied, so add their indices: $x^2x^3=x^5$ . Multiplication of indices belongs to $(x^2)^3$ .

For your turn, vary the inner rational index while keeping the same two-stage method.

Fractional and negative indices

Since $2^4=16$ , the fourth root of $16$ is $2$ . Thus $16^{3/4}$ means $2^3=8$ : the denominator asks for a root and the numerator asks for a power. Taking the root first usually keeps the arithmetic small.

Roots, powers and reciprocals

For $a>0$ , integers $m$ , $n$ with $n>0$ , and rational $r$ ,

a^{m/n}=(\sqrt[n]{a})^m=\sqrt[n]{a^m},\qquad a^{-r}=\frac1{a^r},\qquad a^0=1

Not in the formula booklet: learn it.

The root-first and power-first routes agree. For instance, $27^{2/3}=3^2=9$ , also $\sqrt[3]{27^2}=\sqrt[3]{729}=9$ . Choose the smaller calculation.

Worked example 3: a fractional power

Question. Evaluate $81^{3/4}$ exactly.

Plan. Take the fourth root, then cube.

\begin{aligned} 81^{3/4}&=(\sqrt[4]{81})^3 &&\text{root first}\\ &=3^3=27 &&\text{since }3^4=81 \end{aligned}

Why this step? Taking the fourth root first avoids calculating $81^3$ .

Answer. $27$ .

Check. $27^4=81^3=531441$ , as the power-first interpretation requires.

Change only the sign of the index. This adds a reciprocal instruction; it does not make the answer negative.

Worked example 4: a negative fractional power

Question. Evaluate $81^{-3/4}$ exactly.

Plan. Find the positive power, then take its reciprocal.

\begin{aligned} 81^{-3/4}&=\frac1{81^{3/4}} &&\text{negative index}\\ &=\frac1{3^3}=\frac1{27} &&\text{fourth root, then cube} \end{aligned}

Why this step? A reciprocal reverses multiplication: $81^{3/4}81^{-3/4}=81^0=1$ .

Answer. $1/27$ .

Check. $27\times(1/27)=1$ .

The opening puzzle now makes sense: $(1/16)^{-3/4}=16^{3/4}=8$ . Surprisingly, a small base and a negative exponent can give a larger answer than either number suggests. Similarly, $32^{-2/5}=1/4$ .

Watch out

Wrong: $a^{-r}=-a^r$ . Right: $a^{-r}=1/a^r$ . For a positive base, every rational power remains positive.

Keep exact fractions in this calculation; a rounded decimal would hide the root structure.

Simplifying with common bases

You cannot add the indices in $8^{2/3}16^{1/2}$ immediately: the bases differ. Rewrite $8=2^3$ and $16=2^4$ ; the expression then becomes $2^2\times2^2=16$ . A common base exposes the law you need.

Worked example 5: rewriting numerical bases

Question. Simplify $\dfrac{25^{3/2}5^{-1}}{125^{1/3}}$ .

Plan. Write each base as a power of $5$ .

\begin{aligned} \frac{25^{3/2}5^{-1}}{125^{1/3}}&=\frac{(5^2)^{3/2}5^{-1}}{(5^3)^{1/3}} &&\text{common base}\\ &=5^{3-1-1}=5 &&\text{combine indices} \end{aligned}

Why this step? Once every base is $5$ , one exponent calculation handles the product and quotient.

Answer. $5$ .

Check. Direct evaluation gives $(125/5)/5=5$ .

Change only the exponent on $25$ . The bases and the rest of the calculation stay fixed.

Worked example 6: a negative final index

Question. Simplify $\dfrac{25^{1/2}5^{-1}}{125^{1/3}}$ .

Plan. Use base $5$ , then interpret the final index.

\begin{aligned} \frac{25^{1/2}5^{-1}}{125^{1/3}}&=\frac{(5^2)^{1/2}5^{-1}}{(5^3)^{1/3}} &&\text{rewrite}\\ &=5^{1-1-1}=5^{-1}=\frac15 &&\text{combine} \end{aligned}

Why this step? The denominator contributes a subtracted index, even after the numerator has simplified to $1$ .

Answer. $1/5$ .

Check. $(5\times1/5)/5=1/5$ .

A common base is a choice, not a command to merge unrelated numbers.

Watch out

Wrong: $2^3\times5^2=10^5$ . There is no shared base here. Right: $2^3\times5^2=8\times25=200$ .

For a printed answer, show the rewriting that establishes it. Copying the result is not a demonstration.

Algebraic expressions

With several variables, keep a separate index calculation for each base. Numerical coefficients simplify as ordinary fractions. A product such as $x^{1/2}y^{1/2}$ may be written as $(xy)^{1/2}$ for positive $x$ , $y$ , but unlike bases must not be combined by adding indices.

Worked example 7: two bases

Question. Simplify $\dfrac{6x^{5/2}y^{-1/3}}{3x^{-1/2}y^{2/3}}$ , where $x>0$ , $y>0$ , using positive indices.

Plan. Divide coefficients and subtract indices separately for $x$ and $y$ .

\begin{aligned} \frac{6x^{5/2}y^{-1/3}}{3x^{-1/2}y^{2/3}}&=2x^{5/2-(-1/2)}y^{-1/3-2/3} &&\text{quotient laws}\\ &=2x^3y^{-1}=\frac{2x^3}{y} &&\text{positive indices} \end{aligned}

Why this step? Subtracting a negative denominator index increases the index of $x$ .

Answer. $2x^3/y$ .

Check. At $x=4$ , $y=8$ , the original fraction is $96/6=16$ ; the answer gives $128/8=16$ .

Change only the denominator's index of $y$ . It changes which side of the final fraction contains $y$ .

Worked example 8: subtracting a negative index

Question. Simplify $\dfrac{6x^{5/2}y^{-1/3}}{3x^{-1/2}y^{-4/3}}$ , where $x>0$ , $y>0$ , using positive indices.

Plan. Keep the coefficient and $x$ calculation; recompute the $y$ index.

\begin{aligned} \frac{6x^{5/2}y^{-1/3}}{3x^{-1/2}y^{-4/3}}&=2x^{5/2+1/2}y^{-1/3+4/3} &&\text{subtract negative indices}\\ &=2x^3y &&\text{simplify fractions} \end{aligned}

Why this step? Division by $y^{-4/3}$ multiplies by $y^{4/3}$ .

Answer. $2x^3y$ .

Check. At $x=1$ , $y=8$ , the original is $3/(3/16)=16$ , and $2x^3y=16$ .

This skill returns in differentiation and integration, where a root is often easier to handle as a fractional power.

Exam tip

Show the subtracted indices before simplifying them. That makes a valid method visible even if a fraction slip changes your final power.

Apply the same separate-base calculation with new coefficients and indices.

Quick Recap

A rational index specifies roots and powers; a negative index specifies a reciprocal.
For positive bases, multiply powers by adding indices, divide by subtracting, and raise a power by multiplying.
$a^{m/n}=(\sqrt[n]{a})^m=\sqrt[n]{a^m}$ , and $a^0=1$ for $a\ne0$ .
Rewrite numerical bases when helpful; handle different algebraic bases separately.

The opening puzzle is $(1/16)^{-3/4}=16^{3/4}=8$ : reciprocal, fourth root, cube.

Last Accessed

Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Pure 2.1 - Laws of indices for rational exponents

Meaning of indices

The three core laws

Fractional and negative indices

Simplifying with common bases

Algebraic expressions

Quick Recap

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