Ma12, Ma13 - Rational indices and index laws | OCR A-Level Mathematics B (MEI)

Use Ma12's laws to combine powers and Ma13's negative, fractional and zero indices to interpret what those powers mean. You will keep exact values, make restrictions visible and explain a simplification clearly enough for someone else to follow it.

The three combining laws

Why can $\dfrac{(8x^{-3}y^6)^{2/3}}{2x^{-1}y}$ collapse to such a small expression? First try a concrete case: at $x=2,y=1$ , its numerator and denominator both equal one. We need laws that work for every positive $x,y$ , rather than just this friendly pair.

You already know

$x^3$ means three factors of $x$ . Cancel common nonzero factors in a fraction, and use a common denominator when adding fractional indices.

Counting factors gives $x^3x^2=x^5$ , cancelling gives $x^5/x^2=x^3$ , and two groups of three factors give $(x^3)^2=x^6$ . Rational indices extend that same pattern to roots.

Combining powers

For $x>0$ and rational $a,b$ ,

x^a x^b=x^{a+b},\qquad x^a\div x^b=x^{a-b},\qquad (x^a)^b=x^{ab}.

The first two laws require the same base. The positive-base condition avoids ambiguities with rational powers.

Not in the formula sheet: learn it.

Work from brackets outwards. When a bracket contains a coefficient, that coefficient is also raised to the outside power.

Worked example 1: a coefficient inside the power

Question. For $x>0$ , simplify $(2x^{3/4})^2x^{-1/2}$ .

Plan. Square both factors in the bracket, then collect the powers of $x$ .

\begin{aligned} (2x^{3/4})^2x^{-1/2} &=4x^{3/2}x^{-1/2} && \text{square both factors}\\ &=4x^{3/2-1/2} && \text{add indices}\\ &=4x && \text{simplify} \end{aligned}

Why this step? The outside square acts on $2$ as well as on the power of $x$ .

Answer. $4x$ .

Check. At $x=16$ , the bracket is $16$ , so its square times $16^{-1/2}$ is $256/4=64=4(16)$ .

Change only multiplication by the last factor to division. Subtract its index, including its minus sign.

Worked example 2: subtracting a negative index

Question. For $x>0$ , simplify $\dfrac{(2x^{3/4})^2}{x^{-1/2}}$ .

Plan. Square the bracket, then apply the quotient law.

\begin{aligned} \frac{(2x^{3/4})^2}{x^{-1/2}} &=4x^{3/2-(-1/2)} && \text{square, then subtract}\\ &=4x^2 && \text{combine the fractions} \end{aligned}

Why this step? Dividing by $x^{-1/2}$ is multiplying by $x^{1/2}$ , so the exponent increases.

Answer. $4x^2$ .

Check. At $x=16$ , $256/(1/4)=1024=4(16)^2$ .

A useful numerical surprise comes from the same laws: ten doublings give $2^{10}=1024>10^3$ , so twenty give $2^{20}=1048576>10^6$ . Doubling repeatedly soon overtakes familiar decimal milestones.

Watch out

Wrong: $x^a+x^b=x^{a+b}$ . Addition of indices applies to a product, not a sum. For instance, $2^3+2^2=12$ , whereas $2^{3+2}=32$ . Nor may you combine different bases by adding their indices.

The next divisor changes just the fractional index. Keep the subtraction visible.

Zero powers and reciprocals

A zero index and a negative index describe division, not a special new kind of number.

For $x\ne0$ , $x^a/x^a=1$ , while the quotient law gives $x^{a-a}=x^0$ . Then $x^a x^{-a}=x^0=1$ forces the negative power to be the reciprocal.

Zero and negative indices

For a nonzero base, with all real powers involved defined,

x^0=1,\qquad x^{-a}=\frac1{x^a}.

The first rule does not define $0^0$ ; the second cannot define a negative power of zero.

Not in the formula sheet: learn it.

Invert the whole base before evaluating. This is especially helpful when the base is itself a fraction.

Worked example 3: an exact reciprocal

Question. Evaluate $\left(\frac23\right)^{-2}$ exactly.

Plan. Invert the base, then square numerator and denominator.

\begin{aligned} \left(\frac23\right)^{-2} &=\left(\frac32\right)^2 && \text{negative means reciprocal}\\ &=\frac94 && \text{square both numbers} \end{aligned}

Why this step? The reciprocal reverses the whole fraction; it does not put a minus sign before the answer.

Answer. $\frac94$ .

Check. $(2/3)^2\times9/4=(4/9)(9/4)=1$ .

Now change only the sign of the base. Brackets keep that negative sign inside the square.

Worked example 4: a negative base in brackets

Question. Evaluate $\left(-\frac23\right)^{-2}$ exactly.

Plan. Invert the bracketed fraction, then square it.

\begin{aligned} \left(-\frac23\right)^{-2} &=\left(-\frac32\right)^2 && \text{invert the whole base}\\ &=\frac94 && \text{two negative factors} \end{aligned}

Why this step? An even integer power of a negative number is positive. That is separate from the reciprocal caused by a negative index.

Answer. $\frac94$ .

Check. $(-2/3)^2=4/9$ , and its reciprocal is $9/4$ .

Compare $-4^2=-16$ with $(-4)^2=16$ : in the first expression the minus sign is outside the power. A calculator can check these values, but brackets decide their mathematical meaning.

Watch out

Wrong: $4^{-2}=-16$ . Right: $4^{-2}=1/16$ . Also keep exclusions after cancellation: $x^2/x^2=1$ requires $x\ne0$ , although the final expression looks unrestricted.

Use the same reciprocal-first route with a new negative base.

Roots, rational powers and domains

Read a fractional index as instructions: take the denominator's root, then the numerator's power.

For example, $16^{1/2}=4$ , because $4^2=16$ . The root symbol selects the non-negative answer; it is not asking you to solve an equation with two solutions.

Principal square root

For $u\ge0$ , $\sqrt u$ is the non-negative number whose square is $u$ . Thus $\sqrt9=3$ , while $z^2=9$ has the two solutions $z=\pm3$ . For real $z$ , this gives $\sqrt{z^2}=|z|$ , rather than always $z$ .

For positive bases, roots fit the index laws consistently. Reduce the index first so its root has the intended meaning.

Rational powers

For $x>0$ , integer $m$ and positive integer $n$ ,

x^{m/n}=(\sqrt[n]{x})^m.

A negative numerator adds a reciprocal. Positive powers are also defined at zero; zero and negative powers exclude it.

Not in the formula sheet: learn it.

Taking a root first usually keeps the arithmetic small.

Worked example 5: a negative rational index

Question. Evaluate $64^{-2/3}$ exactly.

Plan. Take the cube root, square it, then take its reciprocal.

\begin{aligned} 64^{-2/3}&=\frac1{(\sqrt[3]{64})^2} && \text{root, power, reciprocal}\\ &=\frac1{4^2} && 4^3=64\\ &=\frac1{16} && \text{evaluate} \end{aligned}

Why this step? The denominator of the index chooses the cube root, while the minus sign applies to the whole resulting power.

Answer. $\frac1{16}$ .

Check. The reciprocal is $16$ , and $16^3=4096=64^2$ .

Change only the numerator of the index, from minus two to minus one. The root stays, but the square disappears.

Worked example 6: changing the numerator

Question. Evaluate $64^{-1/3}$ exactly.

Plan. Take the same cube root and invert it.

\begin{aligned} 64^{-1/3}&=\frac1{\sqrt[3]{64}} && \text{negative fractional index}\\ &=\frac14 && 4^3=64 \end{aligned}

Why this step? The numerator is now one, so there is no further power after the cube root.

Answer. $\frac14$ .

Check. $(1/4)^3=1/64$ .

Odd roots admit negative inputs: $(-8)^{1/3}=-2$ and $(-8)^{2/3}=4$ , with the rational index reduced first. Even roots need non-negative inputs for real answers. Negative bases do not support unrestricted power-of-a-power manipulation: $((-1)^2)^{1/2}=1$ , but $(-1)^{2\times1/2}=-1$ .

Watch out

Wrong: $\sqrt{x^2}=x$ for every real $x$ . Right: $\sqrt{x^2}=|x|$ ; at $x=-2$ , both sides of the correct identity are $2$ . The principal-root identity is not in the formula sheet: learn it.

Keep the domain in view when an exact power is written in a different notation.

Combining laws and comparing a common base

A complicated power becomes manageable when you give its coefficient and each variable their own short calculation.

For positive $u,v$ , $(uv)^r=u^rv^r$ , with rational $r$ . This follows by applying the root and power to both positive factors; it is not in the formula sheet: learn it. In particular, $(2x^3)^2=4x^6$ , not $2x^6$ .

Worked example 7: unpacking the opening puzzle

Question. For $x,y>0$ , simplify $\dfrac{(8x^{-3}y^6)^{2/3}}{2x^{-1}y}$ .

Plan. Apply the outer power to each factor, then subtract each divisor index.

\begin{aligned} \frac{(8x^{-3}y^6)^{2/3}}{2x^{-1}y} &=\frac{4x^{-2}y^4}{2x^{-1}y} && \text{power of each factor}\\ &=2x^{-1}y^3 && \text{quotient laws}\\ &=\frac{2y^3}{x} && \text{positive indices} \end{aligned}

Why this step? The outside $2/3$ multiplies both variable indices and also changes the numerical coefficient.

Answer. $2y^3/x$ , throughout the stated positive domain.

Check. At $x=2,y=1$ , this gives $1$ , matching the opening numerical case.

Change only the outer power. Keeping the other factors fixed makes the effect of that change visible.

Worked example 8: a different outer power

Question. For $x,y>0$ , simplify $\dfrac{(8x^{-3}y^6)^{1/3}}{2x^{-1}y}$ .

Plan. Take the cube root of each factor, then simplify the quotient.

\begin{aligned} \frac{(8x^{-3}y^6)^{1/3}}{2x^{-1}y} &=\frac{2x^{-1}y^2}{2x^{-1}y} && \text{cube root of each factor}\\ &=y && \text{cancel nonzero factors} \end{aligned}

Why this step? Changing $2/3$ to $1/3$ removes the squaring after the cube root; every numerator factor changes accordingly.

Answer. $y$ .

Check. At $x=1,y=2$ , the starting expression is $512^{1/3}/4=8/4=2$ .

The laws also let you compare a common base in an equation:

\begin{aligned} 8^{2x-1}=4^{x+2} &\ \Longleftrightarrow\ 2^{6x-3}=2^{2x+4} && \text{rewrite in base }2\\ &\ \Longleftrightarrow\ 6x-3=2x+4 && \text{equal powers of }2\\ &\ \Longleftrightarrow\ x=\frac74 && \text{solve} \end{aligned}

The two indices both become $15/2$ . Equal powers of a positive base other than one have equal indices; this principle is not in the formula sheet: learn it. It fails for base one, since $1^u=1$ for all real $u$ . Comparing common bases is a useful bridge to exponential equations later.

Exam tip

MEI's “In this question you must show detailed reasoning” asks for the algebra that justifies an exact result. Show the transformed powers or a fraction multiplied out; a decimal check or a calculator's final answer does not supply those steps.

One final change to the outer power returns you to integer-index arithmetic.

Quick Recap

Ma12 combines like bases: add indices in products, subtract in quotients, multiply for a power of a power.
Ma13 interprets zero indices as one, negative indices as reciprocals and fractional indices as roots and powers.
Keep the nonzero restrictions, and use positive bases for the unrestricted rational-index laws.
Exact algebra justifies a result; a numerical check helps you catch slips.

The opening expression is $2y^3/x$ , so its value at $x=2,y=1$ really is one.

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Ma12, Ma13 - Rational indices and index laws

The three combining laws

Zero powers and reciprocals

Roots, rational powers and domains

Combining laws and comparing a common base

Quick Recap

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