2.18 - Index laws and index equations

Use negative and fractional indices as exact instructions, simplify expressions to single powers and solve equations involving powers. You will also recognise when an index equation hides an ordinary quadratic, so that a familiar method becomes available.

Negative and fractional indices

What positive number satisfies $\sqrt x-\frac{12}{\sqrt x}=1$ ? The root and its reciprocal look awkward together. Keep this question in mind: a change of variable will turn it into a quadratic by the final part.

First, make the indices themselves meaningful. A positive whole-number index counts repeated factors: $2^3=2\times2\times2=8$ . A negative index means a reciprocal, not a negative answer. A fractional index introduces a root, not division by the index.

Negative and fractional indices

For a positive base, a negative index takes the reciprocal of the corresponding positive power. In a fractional index $\frac mn$ , the denominator $n$ gives the root and the numerator $m$ gives the power. For example, $16^{1/4}=2$ and $16^{3/4}=2^3=8$ ; $16^{-3/4}$ is the reciprocal of $8$ , not $-8$ .

Taking a larger exponent need not make the value larger. For the base $\frac12$ , $\left(\frac12\right)^2=\frac14$ but $\left(\frac12\right)^3=\frac18$ . A base between zero and one shrinks as the exponent increases. You will use exactly this behaviour in exponential graphs.

Zero, negative and fractional powers

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

\begin{aligned} a^0&=1\\ a^{-m}&=\frac1{a^m}\\ a^{m/n}&=\left(\sqrt[n]{a}\right)^m=\sqrt[n]{a^m} \end{aligned}

The positive-base condition makes all these real powers unambiguous. Zero is not allowed as the base of a negative power; $0^0$ is not defined here. For positive integer powers, negative bases are allowed: use brackets, for example $(-2)^2=4$ .
Not on the formulae sheet: learn it.

For mental arithmetic, take a root before raising it to a power. It keeps the intermediate numbers small.

Worked example 1: root first, then power

Question. Work out $16^{3/4}$ .

Plan. Take the fourth root of $16$ , then cube the result.

\begin{aligned} 16^{3/4}&=\left(16^{1/4}\right)^3 && \text{denominator gives the root}\\ &=2^3 && 2^4=16\\ &=8 && \text{numerator gives the power} \end{aligned}

Why this step? The denominator $4$ asks for a fourth root; dividing $16$ by $4$ would perform the wrong operation.

Answer. $8$ .

Check. $8^4=4096=16^3$ , so this positive value has the required fractional power relationship.

Worked example 2 changes only the sign of the index. The positive power is unchanged inside a reciprocal.

Worked example 2: a negative fractional index

Question. Work out $16^{-3/4}$ .

Plan. Evaluate the corresponding positive power, then take its reciprocal.

\begin{aligned} 16^{-3/4}&=\frac1{16^{3/4}} && \text{negative means reciprocal}\\ &=\frac1{2^3} && \text{fourth root, then cube}\\ &=\frac18 && \text{simplify} \end{aligned}

Why this step? The minus sign belongs to the index. It changes a power into its reciprocal, not its negative.

Answer. $\frac18$ .

Check. $16^{3/4}\times16^{-3/4}=8\times\frac18=1$ , as reciprocal values should.

Zero indices also come from preserving the laws: $\frac{a^3}{a^3}=a^{3-3}=a^0$ , but the fraction is $1$ for $a\ne0$ . A zero exponent does not mean a zero value.

Watch out

Wrong: $16^{-3/4}=-8$ , or $16^{3/4}=\frac{16\times3}{4}$ . Right: use the fourth root, the cube and, for a negative index, the reciprocal.

Use the same sequence with a different fourth power.

Combining powers and roots

Rewrite roots as fractional powers, then apply the same index laws to positive, negative and fractional exponents.

Multiplication joins repeated factors: $3^2\times3^3=3^5$ . Raising a power to another power repeats the whole group: $(3^2)^3=3^6$ . These are different operations, so their effects on the exponents differ.

Index laws

For positive $a$ and $b$ , and rational $p$ and $q$ ,

\begin{aligned} a^p a^q&=a^{p+q}\\ \frac{a^p}{a^q}&=a^{p-q}\\ (a^p)^q&=a^{pq}\\ (ab)^p&=a^p b^p\\ \left(\frac ab\right)^p&=\frac{a^p}{b^p} \end{aligned}

Adding and subtracting indices requires the same base. The positive-base conditions avoid undefined denominators and real-root ambiguities.
Not on the formulae sheet: learn it.

Coefficients obey these rules too: $(2x^3)^2=4x^6$ , not $2x^6$ . A quotient can produce a negative power: $\frac{x^3}{x^5}=x^{-2}=\frac1{x^2}$ , for $x\ne0$ .

Worked example 3: a root over a product

Question. Express $\sqrt{x^{1/2}\times x^{5/2}}$ as a single power of $x$ , where $x>0$ .

Plan. Add the indices inside the root, then multiply the resulting exponent by $\frac12$ .

\begin{aligned} \sqrt{x^{1/2}\times x^{5/2}} &=\sqrt{x^{1/2+5/2}} && \text{multiply powers: add indices}\\ &=(x^3)^{1/2} && \text{a square root is power }\frac12\\ &=x^{3/2} && \text{power of a power: multiply} \end{aligned}

Why this step? The outer root acts on the entire product, so it halves the combined exponent.

Answer. $x^{3/2}$ .

Check. At $x=4$ , the original is $\sqrt{2\times32}=8$ , and $4^{3/2}=8$ .

Worked example 4 changes only the expression inside the root by dividing it by $x^2$ . Subtract that index before applying the outer root.

Worked example 4: a root over a quotient

Question. Express $\sqrt{\frac{x^{1/2}\times x^{5/2}}{x^2}}$ as a single power of $x$ , where $x>0$ .

Plan. Add the numerator indices, subtract the denominator index and finally halve the result.

\begin{aligned} \sqrt{\frac{x^{1/2}\times x^{5/2}}{x^2}} &=\sqrt{x^{1/2+5/2-2}} && \text{combine the inner powers}\\ &=\sqrt{x^1} && \frac12+\frac52-2=1\\ &=x^{1/2} && \text{apply the square root} \end{aligned}

Why this step? The denominator is inside the root, so its exponent is subtracted before the entire result is halved.

Answer. $x^{1/2}$ .

Check. At $x=4$ , the original is $\sqrt{\frac{2\times32}{16}}=2$ , agreeing with $4^{1/2}=2$ .

The scope of the root bar matters as much as a pair of brackets. Work from the inside out.

Watch out

Wrong: $(x^3)^{1/2}=x^{3+1/2}$ . Right: $(x^3)^{1/2}=x^{3/2}$ . Add exponents for a product of powers; multiply them for a power of a power.

Combine a new pair of fractional indices before taking the root.

Solving equations with a fractional power

Translate the index into a root or reciprocal, then undo those operations in a sensible order.

An even root denotes its non-negative value. In $x^{1/2}=4$ , the positive root of $x$ is $4$ , so $x=16$ ; there is no additional answer $x=-16$ . This differs from $x^2=16$ , whose solutions are $x=4$ and $x=-4$ . These inverse-power methods are not supplied as a procedure on the formulae sheet: learn them.

Worked example 5: undoing a square root

Question. Solve $x^{1/2}=4$ .

Plan. Write the fractional power as a square root and square both sides.

\begin{aligned} \sqrt x&=4 && \text{interpret the index}\\ x&=4^2 && \text{square both sides}\\ x&=16 && \text{evaluate} \end{aligned}

Why this step? Both sides are non-negative, so squaring recovers exactly the required value of $x$ .

Answer. $x=16$ .

Check. $\sqrt{16}=4$ , as required.

Worked example 6 changes only the sign of the index. Undo the reciprocal before squaring.

Worked example 6: undoing a reciprocal root

Question. Solve $x^{-1/2}=4$ , where $x>0$ .

Plan. Write the reciprocal root, take reciprocals of both positive sides, then square.

\begin{aligned} \frac1{\sqrt x}&=4 && \text{interpret the negative index}\\ \sqrt x&=\frac14 && \text{take reciprocals}\\ x&=\frac1{16} && \text{square both sides} \end{aligned}

Why this step? The reciprocal of $4$ is $\frac14$ , so the square root is small; it is not $-4$ .

Answer. $x=\frac1{16}$ .

Check. $\left(\frac1{16}\right)^{-1/2}=\frac1{1/4}=4$ .

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2.18 - Index laws and index equations

Negative and fractional indices

Combining powers and roots

Solving equations with a fractional power

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