1.10 Rational exponentsIB Maths: Applications and Interpretation HL: Revision notes

Section 1

What a rational exponent means

A rational exponent is an index that is a fraction. The denominator is a root and the numerator is a power: $a^{\frac1n}=\sqrt[n]{a},\qquad a^{\frac mn}=\left(\sqrt[n]{a}\right)^{m}=\sqrt[n]{a^{m}}.$ So $32^{\frac35}=\left(\sqrt[5]{32}\right)^{3}=2^{3}=8$ and $8^{\frac23}=\left(\sqrt[3]{8}\right)^{2}=4$ . Take the root first because it keeps the numbers small. A negative index means a reciprocal: $a^{-n}=\frac{1}{a^{n}}$ , so $x^{-\frac12}=\frac{1}{\sqrt x}$ and $4^{-\frac12}=\frac12$ .

Key termsrational exponentrootreciprocal

Common mistake

Reading $8^{\frac23}$ as $8\times\frac23$ . The fraction is an index, not a multiplier.

Common mistake

Thinking a negative index makes the value negative. $4^{-\frac12}=\frac12$ , not $-\frac12$ .

Section 2

Laws of exponents with fractional indices

The same laws work for all rational indices: $a^{m}\times a^{n}=a^{m+n},\qquad\frac{a^{m}}{a^{n}}=a^{m-n},\qquad\left(a^{m}\right)^{n}=a^{mn}.$ Add or subtract the fractions in the indices using a common denominator:

$5^{\frac12}\times5^{\frac13}=5^{\frac12+\frac13}=5^{\frac56}$
$6^{\frac34}\div6^{\frac12}=6^{\frac34-\frac12}=6^{\frac14}$
$\left(a^{\frac23}\right)^{\frac32}=a^{1}=a$ The bases must be equal before you combine the indices.

Key termsindex laws

Common mistake

Multiplying the fractions when you should add: $5^{\frac12}\times5^{\frac13}$ is $5^{\frac56}$ , not $5^{\frac16}$ .

Section 3

Simplifying numerically

Write the base as a power of a smaller number, then use the index laws.

$32^{\frac35}=\left(2^{5}\right)^{\frac35}=2^{3}=8$
$8^{\frac23}\times4^{-\frac12}=2^{2}\times2^{-1}=2$
$27^{-\frac23}=\frac{1}{\left(\sqrt[3]{27}\right)^{2}}=\frac19$ Check the result with your GDC by entering the fractional index in brackets, for example $32^{(3/5)}$ .

Key termssimplify

Exam tip

Put every fractional index in brackets on the GDC, or the calculator will divide the wrong part.

Section 4

Simplifying algebraically

Convert roots to indices and then combine like bases. Example: for $x>0$ , $\frac{x^{\frac32}\times x^{-\frac12}}{\sqrt x}=\frac{x^{1}}{x^{\frac12}}=x^{\frac12}$ . Reciprocal roots: $\frac{3}{\sqrt[3]{x^{2}}}=3x^{-\frac23}$ and $\frac{1}{\sqrt x}=x^{-\frac12}$ . Powers of products distribute: $\left(8x^{3}\right)^{\frac13}=2x$ . Always state the final answer with indices in the form $x^{k}$ if asked.

Key termsalgebraic simplification

Section 5

Solving equations and modelling

To solve $x^{\frac mn}=c$ , raise both sides to the reciprocal power $\frac nm$ . Example: $x^{\frac23}=9\Rightarrow x=9^{\frac32}=27$ . Equations such as $x^{\frac12}=3x^{-\frac23}$ combine to $x^{\frac76}=3$ , so $x=3^{\frac67}=2.56$ . Many real formulae have fractional indices. A cube of volume $V$ has edge $s=V^{\frac13}$ and surface area $6V^{\frac23}$ . The orbital period law $T=R^{\frac32}$ rearranges to $R=T^{\frac23}$ . Check units and give the answer to 3 significant figures unless an exact form is requested.

Key termsreciprocal power

Exam tip

Undo the index with its reciprocal: $\frac23$ is undone by $\frac32$ .

That's the notes covered.

Carry on to the next subtopic.

Exam questions on 1.10 Rational exponents

Let $p=8^{\frac23}$ and $q=4^{-\frac12}$ .
Write $pq$ as a single power of $2$ .2 marks
A cube has volume $V$ cm $^3$ and edge length $s$ cm, so that $s=V^{\frac13}$ .
A cube has surface area $150$ cm $^2$ . Find its volume.2 marks
For $x>0$ , consider $A=\frac{x^{\frac32}\times x^{-\frac12}}{\sqrt{x}}$ and $B=\frac{3}{\sqrt[3]{x^{2}}}$ .
Write $A$ in the form $x^{k}$ , where $k\in\mathbb{Q}$ .3 marks

See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).