1B - Integer and fractional indices

Powers do more than count repeated multiplication: they can also describe reciprocals and roots. You will use the index laws to simplify exact expressions and solve equations by writing both sides as powers of the same base.

Products and powers

How can $16^{-\frac34}$ be an exact positive number when its index is negative and fractional?

Start with whole-number powers. In $2^3=2\times2\times2=8$ , the base is $2$ and the index, or exponent, is $3$ . It counts the factors. Putting products together joins their factors:

2^3\times2^2=(2\times2\times2)(2\times2)=2^5.

Dividing cancels factors, while raising a power to another power repeats a whole group. For instance, $(2^3)^2=2^3\times2^3=2^6$ .

Laws of indices

For positive integer indices $m,n$ ,

\begin{aligned} a^m a^n&=a^{m+n}\\ \frac{a^m}{a^n}&=a^{m-n}\qquad(a\ne0)\\ (a^m)^n&=a^{mn}\\ (ab)^n&=a^n b^n \end{aligned}

The quotient rule here gives an ordinary positive power when $m>n$ . The next part explains what happens when $m\le n$ .

Not on the formulae sheet: learn these laws.

Notice the different operations on the indices: addition for a product of powers with the same base, subtraction for division and multiplication for a power of a power. A coefficient inside brackets is raised to the power as well.

Worked example 1: powering a product

Question. Simplify $(3x^2)^3\times x^4$ .

Plan. Cube both factors inside the bracket, then combine the powers of $x$ .

\begin{aligned} (3x^2)^3\times x^4&=3^3(x^2)^3\times x^4 &&\text{cube both factors}\\ &=27x^6\times x^4 &&\text{multiply the nested indices}\\ &=27x^{10} &&\text{add indices for multiplication} \end{aligned}

Answer. $27x^{10}$ .

Check. At $x=2$ , the original is $12^3\times16=27648$ ; the simplified expression is $27\times2^{10}=27648$ . The factor-counting laws establish the result for every $x$ , while the substitution checks the arithmetic.

Change the multiplication by $x^4$ to division. The bracket calculation stays the same, but the final indices must now be subtracted.

Worked example 2: dividing powers

Question. Simplify $\frac{(3x^2)^3}{x^4}$ , where $x\ne0$ .

Plan. Simplify the numerator first, then cancel powers of the same nonzero base.

\begin{aligned} \frac{(3x^2)^3}{x^4}&=\frac{27x^6}{x^4} &&\text{cube each factor}\\ &=27x^{6-4} &&\text{subtract indices}\\ &=27x^2 &&\text{simplify the index} \end{aligned}

Answer. $27x^2$ , for $x\ne0$ .

Check. At $x=2$ , the original is $12^3\div16=108$ ; the answer is $27\times4=108$ . The original fraction is undefined at $x=0$ , so that restriction stays even though the final expression can be evaluated there.

The laws depend on the structure of the expression, not just on seeing powers.

Watch out

$a^m+a^n$ does not become $a^{m+n}$ . For example, $2^3+2^2=12$ , whereas $2^5=32$ . Also, $(3x^2)^3=27x^6$ , not $3x^6$ : the coefficient is inside the cubed bracket.

Use the quotient structure in the next expression.

Zero and negative indices

A zero or negative index keeps the quotient law consistent when there are no factors left in the numerator.

For example, $5^3\div5^3=1$ , but subtracting indices gives $5^{3-3}=5^0$ . Therefore $5^0=1$ . Dividing one more time by $5$ gives $5^{-1}=\frac15$ , and again gives $5^{-2}=\frac1{25}$ . Each step reduces the index by one and divides the value by $5$ .

Zero and negative indices

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

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

These definitions extend the index laws to all integer indices. The expression $0^0$ is not defined by this rule, and a negative power of zero is undefined because it would involve division by zero.

Not on the formulae sheet: learn it.

The minus sign in the index means “take the reciprocal”. It does not make the value negative.

Worked example 3: a negative index

Question. Find the exact value of $5^{-2}$ .

Plan. Replace the negative power with the reciprocal of a positive power.

\begin{aligned} 5^{-2}&=\frac{1}{5^2} &&\text{take the reciprocal}\\ &=\frac1{25} &&\text{evaluate }5^2 \end{aligned}

Answer. $\frac1{25}$ .

Check. $\frac1{25}\times5^2=1$ , as required by $5^{-2}5^2=5^0$ .

Now replace the base by its reciprocal. Taking a reciprocal twice changes a small positive result into a large positive result.

Worked example 4: a fractional base

Question. Find the exact value of $\left(\frac15\right)^{-2}$ .

Plan. Take the reciprocal of the whole squared fraction.

\begin{aligned} \left(\frac15\right)^{-2} &=\frac{1}{\left(\frac15\right)^2} &&\text{use the negative index}\\ &=\frac{1}{\frac1{25}} &&\text{square numerator and denominator}\\ &=25 &&\text{divide by a fraction} \end{aligned}

Answer. $25$ .

Check. The original base is $5^{-1}$ , so $(5^{-1})^{-2}=5^2=25$ by the power-of-a-power law.

For a general nonzero fraction, the same idea gives $\left(\frac ab\right)^{-n}=\left(\frac ba\right)^n$ , with $a\ne0$ and $b\ne0$ . You can turn the fraction over first, then apply the positive power.

Watch out

$5^{-2}\ne-25$ . A negative index describes a reciprocal. A negative base is a different matter: $(-5)^2=25$ , while $-5^2=-(5^2)=-25$ because the minus sign is outside the power.

Keep the whole fraction inside the brackets when finding the next value.

Fractional indices

A fractional index describes a root because it must obey the power-of-a-power law.

If $u=16^{\frac12}$ , then $u^2=16^{\frac12\times2}=16$ . For the positive base $16$ , we choose the positive value $u=4$ . Thus $16^{\frac12}=\sqrt{16}$ . In the same way, $8^{\frac13}=2$ , because $2^3=8$ .

A rational power

For a positive base $a$ , the power $a^{\frac mn}$ uses an integer numerator $m$ and positive integer denominator $n$ . The denominator gives the root and the numerator gives the power.

Use a positive base in these rules so the root and all equivalent fractional forms have one consistent positive real value.

You may take the root first or the power first; taking the root first usually keeps the arithmetic smaller.

Fractional indices

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

a^{\frac1n}=\sqrt[n]{a},\qquad a^{\frac mn}=(\sqrt[n]{a})^m=\sqrt[n]{a^m}.

For a negative fractional index, take a reciprocal as well:

a^{-\frac mn}=\frac1{a^{\frac mn}}.

The index laws continue to apply to these rational powers of positive bases.

Not on the formulae sheet: learn it.

These powers connect directly to the surds you already know. For example, $12^{\frac12}=\sqrt{12}=2\sqrt{3}$ ; a fractional power need not be an integer.

Worked example 5: a root followed by a power

Question. Find the exact value of $16^{\frac34}$ .

Plan. Take the fourth root first, then cube it.

\begin{aligned} 16^{\frac34}&=(\sqrt[4]{16})^3 &&\text{denominator }4\text{ gives the root}\\ &=2^3 &&\text{since }2^4=16\\ &=8 &&\text{numerator }3\text{ gives the power} \end{aligned}

Answer. $8$ .

Check. Raising the answer to the fourth power gives $8^4=4096=16^3$ , as $(16^{\frac34})^4=16^3$ requires. Both values are positive.

Now change only the sign of the index. The root and cube are unchanged; the result must be inverted.

Worked example 6: a negative fractional index

Question. Find the exact value of $16^{-\frac34}$ .

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

\begin{aligned} 16^{-\frac34}&=\frac1{16^{\frac34}} &&\text{negative means reciprocal}\\ &=\frac1{2^3} &&\text{take the fourth root first}\\ &=\frac18 &&\text{evaluate the cube} \end{aligned}

Answer. $\frac18$ , answering the opening question.

Check. $\left(\frac18\right)^4=\frac1{4096}=16^{-3}$ , and the answer is positive, as a real power of a positive base must be.

The root sign and an equation asking for roots still mean different things: $16^{\frac12}=4$ , while $x^2=16$ has solutions $x=4$ and $x=-4$ .

Watch out

$16^{\frac34}\ne\frac{16^3}{4}$ . The denominator of the index specifies a fourth root; it is not a divisor outside the power. Read the index before calculating.

Use the same root–power–reciprocal decisions with a cube root.

Equations with a common base

You can solve some equations with an unknown index by rewriting both sides using one base.

For example, $4^x=8$ becomes $(2^2)^x=2^3$ , so $2^{2x}=2^3$ . Equal powers of the same positive base other than $1$ have equal exponents, giving $2x=3$ .

Why can we equate the exponents? For a base greater than $1$ , increasing the exponent increases the value; for a base between $0$ and $1$ , increasing the exponent decreases the value. The same value is reached only once. You will see both shapes in the next lesson.

Equating exponents

For $a>0$ and $a\ne1$ ,

a^u=a^v\quad\Longrightarrow\quad u=v.

Rewrite both sides with the same base before using this result. Base $1$ is excluded because $1^u=1^v$ for all real $u,v$ .

Not on the formulae sheet: learn it.

Choose a base that makes the given numbers exact powers. Reciprocals and roots can supply negative or fractional indices.

Worked example 7: two whole-number bases

Question. Solve $4^{x+1}=8^x$ .

Plan. Write both bases as powers of $2$ , then equate the exponents.

\begin{aligned} (2^2)^{x+1}&=(2^3)^x &&\text{use }4=2^2,\;8=2^3\\ 2^{2x+2}&=2^{3x} &&\text{multiply the indices}\\ 2x+2&=3x &&\text{equate exponents}\\ x&=2 &&\text{solve the linear equation} \end{aligned}

Answer. $x=2$ .

Check. Substitution in the original equation gives $4^3=64$ and $8^2=64$ .

Replace the right-hand base by its reciprocal. This changes the sign of its exponent when you rewrite it using $2$ .

Worked example 8: a reciprocal base

Question. Solve $4^{x+1}=\left(\frac18\right)^x$ .

Plan. Use $4=2^2$ and $\frac18=2^{-3}$ .

\begin{aligned} 2^{2(x+1)}&=2^{-3x} &&\text{rewrite both bases}\\ 2x+2&=-3x &&\text{equate exponents}\\ 5x&=-2 &&\text{collect the }x\text{ terms}\\ x&=-\frac25 &&\text{divide by }5 \end{aligned}

Answer. $x=-\frac25$ .

Check. In the original equation, the left side is $4^{\frac35}=2^{\frac65}$ ; the right side is $(2^{-3})^{-\frac25}=2^{\frac65}$ . The exact values agree.

If no convenient common base is available, do not equate the exponents of different bases. Logarithms will give you a method for those equations later.

Watch out

From $4^{x+1}=8^x$ , writing $x+1=x$ ignores the unequal bases. First rewrite the bases, giving the valid equation $2x+2=3x$ .

Negative and fractional indices are therefore consistent extensions of repeated multiplication: our opening value is $16^{-\frac34}=\frac18$ , and the same laws let us check exact fractional solutions to equations.

Quick Recap

Multiply powers of the same base by adding indices; divide by subtracting them.
A power of a power multiplies indices, and a bracketed coefficient is powered too.
For a nonzero base, a zero index gives $1$ and a negative index gives a reciprocal.
For a positive base, the denominator of a fractional index gives the root; the numerator gives the power.
Equate exponents only after obtaining the same positive base other than $1$ .

The opening expression is $16^{-\frac34}=\frac1{(\sqrt[4]{16})^3}=\frac18$ .

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1B - Integer and fractional indices

Products and powers

Zero and negative indices

Fractional indices

Equations with a common base

Quick Recap

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