E2.4 - Fractional powers with letters | Cambridge IGCSE Mathematics

Fractional powers connect algebraic indices to roots. You will simplify fractional and negative powers with their real-number conditions, combine index laws and solve common-base equations whose exponents contain linear expressions. This lesson is for the Extended route.

Fractional powers of algebraic terms

How can a square root turn into a reciprocal without any new operation? For a positive number $x$ , $x^{1/2}$ is its square root, while $x^{-1/2}=\frac1{\sqrt{x}}$ . The minus sign still means “take the reciprocal”; the fraction still means “take a root”. Each part of the index keeps its job.

For example, $16^{3/4}=(\sqrt[4]{16})^3=2^3=8$ . Take the root first when it makes the numbers simpler. Letters follow the same structure, but their possible values must be checked.

Real fractional power

For an integer $m$ and positive integer $n$ , with $\frac mn$ in lowest terms, define

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

For an even $n$ , a real root requires $x\ge0$ , and $\sqrt[n]{x}$ means the non-negative root. An odd root also allows negative $x$ . A negative power additionally excludes $x=0$ , because it introduces a reciprocal. Zero to a positive rational power is $0$ ; zero to a negative or zero power is not defined here.

Thus $(-8)^{1/3}=-2$ is real, whereas $(-8)^{1/2}$ is not real. Be especially careful with an even root of a square: $\sqrt{x^2}$ is the non-negative size of $x$ , so at $x=-3$ it is $3$ , not $-3$ . In the simplifications below we state positive variables so that the fractional-power laws can be applied directly.

Worked example 1: a root of an algebraic term

Question. Simplify $(16x^8)^{1/4}$ , where $x>0$ .

Plan. Take the fourth root of the coefficient and the power of $x$ .

\begin{aligned} (16x^8)^{1/4}&=16^{1/4}(x^8)^{1/4} && \text{positive factors}\\ &=2x^{8/4} && \text{fourth root and index product}\\ &=2x^2 && \text{simplify the index} \end{aligned}

Answer. $2x^2$ .

Check. $2x^2>0$ and $(2x^2)^4=16x^8$ , so it is the required non-negative fourth root.

Change only the sign of the outside index. The root is the same; now take its reciprocal.

Worked example 2: the reciprocal of the root

Question. Simplify $(16x^8)^{-1/4}$ , where $x>0$ .

Plan. Use the negative-index rule before simplifying the fourth root.

\begin{aligned} (16x^8)^{-1/4}&=\frac1{(16x^8)^{1/4}} && \text{negative index}\\ &=\frac1{2x^2} && \text{evaluate the fourth root} \end{aligned}

Answer. $\frac1{2x^2}$ .

Check. At $x=2$ , the original is $4096^{-1/4}=\frac18$ ; the answer is $\frac1{2\times4}=\frac18$ .

A numerator greater than $1$ adds a power after the root. For $t>0$ , $(27t^6)^{2/3}=(3t^2)^2=9t^4$ . The denominator $3$ chooses the cube root; the numerator $2$ squares it.

Watch out

$x^{1/2}=\frac{x}{2}$ is wrong: at $x=9$ , the square root is $3$ , not $4.5$ . Also, $x^{-1/2}$ means $\frac1{\sqrt{x}}$ , not $-\sqrt{x}$ .

For a negative quarter power, find the fourth root before taking its reciprocal.

Combining fractional index laws

For positive bases, the familiar index laws also work with fractional indices. For instance, $x^{1/2}x^{1/2}=\sqrt{x}\sqrt{x}=x=x^1$ , agreeing with $\frac12+\frac12=1$ . The same-base requirement still matters, and coefficients still use ordinary arithmetic.

Fractional index laws with positive bases

For $x>0$ , $y>0$ and rational $r,s$ ,

\begin{aligned} x^r x^s&=x^{r+s} & \frac{x^r}{x^s}&=x^{r-s}\\ (x^r)^s&=x^{rs} & (xy)^r&=x^r y^r \end{aligned}

The positive-base conditions make every displayed power real and avoid zero denominators. Do not extend the power-of-a-power rule automatically to negative bases: $((-3)^2)^{1/2}=3$ , whereas $(-3)^{2\times1/2}=-3$ . Not given in the List of formulas: learn the laws and their conditions.

Use a common denominator when adding or subtracting fractional indices. For example, $x^{1/3}x^{1/2}=x^{2/6+3/6}=x^{5/6}$ . An outside fractional power multiplies indices: for $x>0$ , $(x^{2/3})^{3/4}=x^{1/2}$ .

Worked example 3: multiplying fractional powers

Question. Simplify $6x^{-1/2}\times\frac13x^{5/2}$ , where $x>0$ .

Plan. Multiply the coefficients and add the indices of $x$ .

\begin{aligned} 6x^{-1/2}\times\frac13x^{5/2}&=2x^{-1/2+5/2} && \text{multiply same-base powers}\\ &=2x^2 && \text{add the fractions} \end{aligned}

Answer. $2x^2$ .

Check. At $x=4$ , the factors are $3$ and $\frac{32}{3}$ , whose product is $32$ . The answer gives $2\times16=32$ .

Change only multiplication to division. The indices must now be subtracted, and the coefficients must be divided.

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E2.4 - Fractional powers with letters

Fractional powers of algebraic terms

Combining fractional index laws

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