1.4 - Powers, roots and index laws

Powers record repeated multiplication, and roots reverse it. You will recognise square and cube numbers, calculate roots and use integer index laws, including zero and negative indices. The final part adds fractional indices for positive numerical bases.

Squares, cubes and roots

Can $64$ be both a square number and a cube number? A square arranges equal rows and columns; a cube has three equal dimensions. Powers let us describe both arrangements compactly.

You already know

Multiplying two negative numbers gives a positive number. Powers are calculated before a leading minus sign, unless brackets include that sign in the base.

For example, $3^2=3\times3=9$ , while $3^3=3\times3\times3=27$ . The index counts factors, not additions.

Powers and roots

In $a^n$ , $a$ is the base and $n$ is the index. For a positive integer $n$ , multiply $n$ copies of the base. A square number is an integer squared; a cube number is an integer cubed. A root reverses a power. The symbol $\sqrt a$ selects the nonnegative square root; $\sqrt[3]{a}$ is the cube root.

The positive squares begin $1, 4, 9, 16, 25, 36, 49, 64, 81, 100$ ; the positive cubes begin $1, 8, 27, 64, 125$ . Zero is also both a square and a cube. These facts and root meanings are not given in either tier's formulae sheet: learn them.

Worked example 1: reverse positive powers

Question. Work out $6^2$ , $6^3$ , $\sqrt{81}$ and $\sqrt[3]{64}$ .

Plan. Multiply for powers; find the number that reproduces the radicand for each root.

\begin{aligned} 6^2&=6\times6=36 &&\text{two equal factors}\\ 6^3&=6\times6\times6=216 &&\text{three equal factors}\\ \sqrt{81}&=9 &&\text{since }9^2=81\\ \sqrt[3]{64}&=4 &&\text{since }4^3=64 \end{aligned}

Why choose $9$ , not $-9$ , for $\sqrt{81}$ ? The square-root symbol means the nonnegative root.

Answer. $36, 216, 9, 4$ , respectively.

Check. $36\div6=6$ , $216\div6\div6=6$ , and raising each root to its indicated power gives its starting number.

Now change the signs of the bases and the cube-root input. Brackets determine whether the minus sign is being raised to a power.

Worked example 2: include negative signs

Question. Work out $(-6)^2$ , $-6^2$ , $(-6)^3$ and $\sqrt[3]{-64}$ .

Plan. Identify each full base, then count negative factors.

\begin{aligned} (-6)^2&=(-6)(-6)=36 &&\text{two negative factors}\\ -6^2&=-(6\times6)=-36 &&\text{power before the leading minus}\\ (-6)^3&=(-6)(-6)(-6)=-216 &&\text{three negative factors}\\ \sqrt[3]{-64}&=-4 &&\text{since }(-4)^3=-64 \end{aligned}

Why can a cube root be negative? Three negative factors have a negative product; two have a positive product.

Answer. $36, -36, -216, -4$ , respectively.

Check. The even power is positive, the odd power is negative, and $(-4)(-4)(-4)=-64$ .

A negative number has no real square root, because every real square is nonnegative. This differs from a negative cube: $(-3)^3=-27$ . Also, $\sqrt{81}=9$ differs from solving $t^2=81$ , whose two solutions are $t=9$ and $t=-9$ .

Watch out

Wrong: $\sqrt{81}=\pm9$ . The symbol selects one value, $9$ . Use both signs when solving $t^2=81$ , not when evaluating the root symbol alone.

The opening puzzle works: $64=8^2=4^3$ . More generally, a sixth power is both a square and a cube. For example, $2^6=(2^3)^2=(2^2)^3=64$ . The next part explains why those indices multiply.

Multiplying and dividing equal bases

Multiplying $3^4$ by $3^2$ joins four factors of $3$ to two more. There are six factors in total, so $3^4\times3^2=3^6$ . Dividing cancels factors instead.

Integer index laws

For a nonzero base $a$ and integer indices $m$ and $n$ ,

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

For nonzero $a$ and $b$ , $(ab)^n=a^n b^n$ . Positive-index products also allow zero bases. These laws are absent from both tier sheets. Not in the formulae sheet: learn it.

In a power of a power, $n$ copies of $a^m$ contain $mn$ copies of $a$ . In a power of a product, each copy contains both factors. For example, $(2\times5)^3=2^3\times5^3=1000$ . The next part defines zero and negative indices so the same laws keep working.

Worked example 3: add and subtract indices

Question. Write $3^4\times3^2\div3^3$ as one power of $3$ , then work out its value.

Plan. Add indices for multiplication and subtract the divisor's index.

\begin{aligned} 3^4\times3^2\div3^3&=3^{4+2-3} &&\text{all bases are }3\\ &=3^3 &&\text{combine the indices}\\ &=27 &&\text{evaluate the power} \end{aligned}

Why subtract $3$ at the division step? Three factors cancel from the six-factor numerator.

Answer. $3^3$ , with value $27$ .

Check. $81\times9\div27=27$ .

Now change the numerator from a product of powers to a power of a power. This changes addition of its indices to multiplication.

Worked example 4: raise a power to a power

Question. Write $(3^4)^2\div3^3$ as one power of $3$ , then work out its value.

Plan. Multiply the bracketed indices first, then subtract the divisor's index.

\begin{aligned} (3^4)^2\div3^3&=3^{4\times2}\div3^3 &&\text{two copies of }3^4\\ &=3^{8-3} &&\text{cancel three factors}\\ &=3^5=243 &&\text{evaluate the remaining power} \end{aligned}

Why multiply $4$ and $2$ ? Each of the two bracketed factors contains four factors of $3$ .

Answer. $3^5$ , with value $243$ .

Check. $81^2\div27=6561\div27=243$ .

Equal bases are essential for adding or subtracting indices. Addition of values is a different operation too: $3^2+3^2=18$ , whereas $3^2\times3^2=81$ .

Watch out

Wrong: $(3^4)^2=3^{4+2}=3^6$ . The bracket is repeated twice, so the correct power is $3^{4\times2}=3^8$ . Add indices for a product of equal bases; multiply them for a power of a power.

If a question asks for one power and its value, give both forms. A lone numerical answer does not supply the requested index notation.

Zero and negative integer indices

A power with index zero does not mean no value. Compare $5^2\div5^2$ : the quotient is $1$ , while the index law gives $5^{2-2}=5^0$ . The two must agree.

Zero and negative indices

For $a\ne0$ and a positive integer $n$ ,

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

A negative index means a reciprocal, not a negative answer. These results are absent from both the Foundation and Higher sheets. Not in the formulae sheet: learn it.

Dividing by the base repeatedly gives $2^2=4$ , $2^1=2$ , $2^0=1$ , $2^{-1}=\frac12$ . This keeps the index laws consistent. The base cannot be zero: taking its reciprocal would mean dividing by zero. The quotient argument does not define $0^0$ .

Worked example 5: evaluate zero and negative indices

Question. Work out $5^0$ and $2^{-3}$ .

Plan. Use $1$ for the zero power of a nonzero base, and take a reciprocal for the negative power.

\begin{aligned} 5^0&=1 &&\text{nonzero base, zero index}\\ 2^{-3}&=\frac1{2^3} &&\text{negative index means reciprocal}\\ &=\frac18 &&\text{cube the denominator} \end{aligned}

Why a reciprocal? $2^3\times2^{-3}=2^0=1$ , so the two values must multiply to $1$ .

Answer. $1$ and $\frac18$ , respectively.

Check. $8\times\frac18=1$ .

Now change only the base of the negative power from $2$ to $-2$ . The negative index still means reciprocal; the base's sign controls the sign of its cube.

Worked example 6: separate the two minus signs

Question. Work out $5^0$ and $(-2)^{-3}$ .

Plan. Keep the bracketed negative base when forming the reciprocal.

\begin{aligned} 5^0&=1 &&\text{same nonzero base rule}\\ (-2)^{-3}&=\frac1{(-2)^3} &&\text{invert the positive-index power}\\ &=\frac1{-8}=-\frac18 &&\text{an odd power keeps the minus sign} \end{aligned}

Why is this answer negative? The denominator is the cube of a negative base, not because the index is negative.

Answer. $1$ and $-\frac18$ , respectively.

Check. $(-8)\times\left(-\frac18\right)=1$ .

The index laws also simplify mixed integer powers. For example, $5^{-2}\times5^3=5^{-2+3}=5$ . A negative index can disappear after you combine powers.

Watch out

Wrong: $2^{-3}=-8$ . Changing the index's sign takes a reciprocal, so $2^{-3}=\frac18$ . The value $-8$ is $(-2)^3$ , which has a different base and index.

Keep the base and index separate when reading a question. This helps with fractional indices next too.

Fractional indices (Higher)

A fractional index combines a root and a power. Start with $64^{1/3}$ : cubing it must give $64$ , because multiplying the indices gives $1$ . Its positive value is therefore the cube root, $4$ .

Fractional indices

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

a^{1/n}=\sqrt[n]{a},\qquad a^{m/n}=\left(\sqrt[n]{a}\right)^m

The denominator gives the root; the numerator gives the power. A negative index also takes a reciprocal. This result is absent from the Higher sheet. Not in the formulae sheet: learn it.

Taking the root first often keeps the numbers small. The bases here are positive, so these root-and-power transformations give unambiguous positive real values. The cube roots of negative numbers from the first part remain valid, but do not assume every fractional-power transformation works for a negative base.

Worked example 7: take the root before the power

Question. Work out $64^{2/3}$ .

Plan. Take the cube root, then square it.

\begin{aligned} 64^{2/3}&=\left(\sqrt[3]{64}\right)^2 &&\text{denominator }3\text{ gives cube root}\\ &=4^2 &&\text{since }4^3=64\\ &=16 &&\text{numerator }2\text{ gives square} \end{aligned}

Why root first? Cubing $4$ is easy to recognise; squaring $64$ first would create a larger number to handle.

Answer. $16$ .

Check. $16^3=4096=64^2$ , as the indices predict.

Now change only the sign of the index. Keep the same root-and-power calculation, then take its reciprocal.

Worked example 8: include a negative fractional index

Question. Work out $64^{-2/3}$ . Give an exact answer.

Plan. Find $64^{2/3}$ , then invert the result.

\begin{aligned} 64^{-2/3}&=\frac1{64^{2/3}} &&\text{negative index means reciprocal}\\ &=\frac1{\left(\sqrt[3]{64}\right)^2} &&\text{use a cube root and square}\\ &=\frac1{4^2}=\frac1{16} &&\text{evaluate exactly} \end{aligned}

Why is the answer still positive? The base is positive; the minus sign in the index changes size, not sign.

Answer. $\frac1{16}$ .

Check. $64^{2/3}\times64^{-2/3}=16\times\frac1{16}=1$ .

Other denominators request other roots: $16^{-3/4}=\frac1{(\sqrt[4]{16})^3}=\frac1{2^3}=\frac18$ . A positive fractional base works too: $(\frac1{25})^{3/2}=(\frac15)^3=\frac1{125}$ . The same index laws apply; for example, $16^{1/4}\times16^{1/2}=16^{3/4}=8$ .

Watch out

Wrong: $64^{2/3}=\sqrt{64^3}$ . That swaps the numerator and denominator. The denominator $3$ asks for a cube root, then the numerator $2$ asks for a square: $(\sqrt[3]{64})^2=16$ .

The opening number $64$ was both a square and a cube. That makes it especially convenient for fractional powers, because both relevant roots are whole numbers.

Quick Recap

A square root symbol selects the nonnegative root; a cube root can be negative.
Add indices when multiplying equal bases, subtract when dividing, and multiply indices for a power of a power.
For nonzero bases, a zero index gives $1$ ; a negative index gives a reciprocal.
Fractional indices use roots and powers, with positive bases in this lesson's final part.

Yes, the opening number is both a square and a cube: $64=8^2=4^3$ .

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1.4 - Powers, roots and index laws

Squares, cubes and roots

Multiplying and dividing equal bases

Zero and negative integer indices

Fractional indices (Higher)

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