Laws of indices for all rational exponents

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Algebra and functions is the largest block of Pure Mathematics P1 (WMA11/01): on each of the three recent P1 papers studied for these notes it carried 32 of the 75 marks.

This first section is about indices, the small raised numbers that say how many times a factor is repeated.

They look like a GCSE topic, and you will have used them in International GCSE Mathematics, but in P1 they extend to negative and fractional powers, and every calculus question later in the course depends on rewriting expressions as powers of $x$ .

The idea: an index counts factors

$a^5$ means $a \times a \times a \times a \times a$ : five factors of $a$ .

Every law of indices is just counting.

$a^3 \times a^4$ is three factors of $a$ followed by four more, so seven factors: $a^7$ .

$a^6 \div a^2$ cancels two of the six factors, leaving four: $a^4$ .

$(a^2)^3$ is three copies of $a^2$ , so $2 + 2 + 2 = 6$ factors: $a^6$ .

Written in general, these are the three laws to learn.

They hold for any base $a$ (with $a \neq 0$ wherever you divide), and the work below shows they can be kept for all rational exponents, not only whole numbers.

Why $a^0 = 1$ and $a^{-n} = \dfrac{1}{a^n}$

Counting factors only makes sense for positive whole-number indices.

Mathematicians extend the meaning by insisting that the laws keep working.

Take $a^3 \div a^3$ . By the division law it is $a^{3-3} = a^0$ . But anything divided by itself is 1. So $a^0 = 1$ is forced on us.

Take $a^0 \div a^n = a^{0-n} = a^{-n}$ . The left-hand side is $\dfrac{1}{a^n}$ . So $a^{-n} = \dfrac{1}{a^n}$ : a negative index means "one over".

Neither result is a new rule to memorise; each is the only value that keeps the laws true.

Why $a^{\frac{1}{n}}$ is an $n\text{th}$ root

Apply the power law to $a^{\frac{1}{2}}$ : $\left(a^{\frac{1}{2}}\right)^2 = a^{\frac{1}{2} \times 2} = a^1 = a$ .

So $a^{\frac{1}{2}}$ is a number whose square is $a$ : it is $\sqrt{a}$ .

In the same way $\left(a^{\frac{1}{3}}\right)^3 = a$ , so $a^{\frac{1}{3}} = \sqrt[3]{a}$ .

Then

a^{\frac{m}{n}} = \left(a^{\frac{1}{n}}\right)^m = \left(\sqrt[n]{a}\right)^m \quad\text{and also}\quad a^{\frac{m}{n}} = \left(a^m\right)^{\frac{1}{n}} = \sqrt[n]{a^m}.

That is the equivalence you are expected to know.

For arithmetic without a calculator, take the root first, because the numbers stay small: $32^{\frac{3}{5}} = \left(\sqrt[5]{32}\right)^3 = 2^3 = 8$ , where $32^3 = 32\,768$ would be an awkward number to find a fifth root of.

A quick numerical check that this is sensible: $4^{\frac{1}{2}} = 2$ sits between $4^0 = 1$ and $4^1 = 4$ , and $2$ is exactly the value that makes $4^0, 4^{\frac{1}{2}}, 4^1$ go up by the same factor each half-step (1, 2, 4).

Rewriting with one base

Many P1 questions give a letter such as $u = 3^x$ and ask for other expressions "in terms of $u$ ".

The method is always the same: write every base as a power of the same prime, then use the laws to split off the part that is $u$ .

$9^x = (3^2)^x = 3^{2x} = (3^x)^2 = u^2$ .

$3^{x+2} = 3^x \times 3^2 = 9u$ . The "+2" in the index becomes a factor of 9, not "+2".

Where this leads

Indices are the language of the calculus in P1: to differentiate or integrate $\dfrac{3}{\sqrt{x}}$ you first write it as $3x^{-\frac{1}{2}}$ .

In P2 the same ideas lead to exponential functions $a^x$ and logarithms, which undo them, and the hidden quadratics later in this unit use substitutions like $u = 2^x$ .

Outside mathematics, fractional powers appear in physical laws: the period of a planet's orbit is proportional to (orbit radius) $^{\frac{3}{2}}$ .

How the marks work

M1: a correct method applied to this question, with your values in. A later slip does not lose it. dM1 is a method mark that needs the M mark before it.

A1: a correct answer or line, which needs its M mark. A1* is the last mark of a "show that": the printed result, reached with no errors and at least one line before it. A1cso: correct solution only.

B1: an independent mark for a correct value, fact or sketch feature. ft (follow through): credit for correct use of your own earlier wrong value.

awrt: any answer which rounds to the value given. isw: ignore subsequent working, so a correct answer followed by a harmless extra line keeps its marks.

Misread: copy a number wrongly but keep the character of the question, and you lose two of the A or B marks in that part, not everything.

Some questions print a note that answers relying on calculator technology are not acceptable. Under it, a correct answer with no working can score nothing.

Key terms

Laws of indices for all rational exponents: $a^m \times a^n = a^{m+n}$ , $a^m \div a^n = a^{m-n}$ , $(a^m)^n = a^{mn}$ .

Fractional index: the equivalence of $a^{\frac{m}{n}}$ and $\sqrt[n]{a^m}$ should be known.

Negative index: $a^{-n} = \dfrac{1}{a^n}$ , for $a \neq 0$ .

Zero index: $a^0 = 1$ , for $a \neq 0$ .

Worked example 1: values without a calculator

Find the exact value of (a) $27^{\frac{2}{3}}$ (b) $16^{-\frac{3}{4}}$ (c) $\left(\dfrac{8}{125}\right)^{-\frac{1}{3}}$ .

(4)

1. (a) $27^{\frac{2}{3}} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9$ B1

2. (b) $16^{-\frac{3}{4}} = \dfrac{1}{16^{\frac{3}{4}}} = \dfrac{1}{\left(\sqrt[4]{16}\right)^3}$ M1 (negative index as a reciprocal, root taken)

3. $= \dfrac{1}{2^3} = \dfrac{1}{8}$ A1

4. (c) $\left(\dfrac{8}{125}\right)^{-\frac{1}{3}} = \left(\dfrac{125}{8}\right)^{\frac{1}{3}} = \dfrac{5}{2}$ B1

Answer: (a) 9 (b) $\dfrac{1}{8}$ (c) $\dfrac{5}{2}$ .

Check (b) by size: $16^{\frac{3}{4}}$ is between $16^{\frac{1}{2}} = 4$ and $16^1 = 16$ , and it is 8, so its reciprocal is $\frac{1}{8}$ .

Worked example 2: the form $kx^n$

Express each of the following in the form $kx^n$ , where $k$ and $n$ are constants in simplest form.

(a) $\sqrt[3]{27x^{-6}}$ (2)

(b) $\dfrac{(2x^3)^4}{8\sqrt{x}}$ (2)

1. (a) $\left(27x^{-6}\right)^{\frac{1}{3}} = 27^{\frac{1}{3}} \times x^{-6 \times \frac{1}{3}}$ M1 (power law on each factor)

2. $= 3x^{-2}$ A1

3. (b) $(2x^3)^4 = 16x^{12}$ and $8\sqrt{x} = 8x^{\frac{1}{2}}$ , so the expression is $\dfrac{16x^{12}}{8x^{\frac{1}{2}}}$ M1

4. $= 2x^{12 - \frac{1}{2}} = 2x^{\frac{23}{2}}$ A1

Answer: (a) $3x^{-2}$ (b) $2x^{\frac{23}{2}}$ .

In (b), the 4 applies to the 2 as well as to $x^3$ : $2^4 = 16$ , a step that is easy to drop.

Worked example 3: a substitution letter, then a show that

Given that $u = 3^x$ ,

(a) express $3^{x+2}$ in terms of $u$ . (1)

(b) Express $27^{x-1}$ in terms of $u$ . (2)

1. (a) $3^{x+2} = 3^x \times 3^2 = 9u$ B1

2. (b) $27^{x-1} = 3^{3(x-1)} = 3^{3x} \times 3^{-3}$ M1 (same prime base, laws used)

3. $= \dfrac{u^3}{27}$ A1

4. (c) $\dfrac{9^{x+1}}{3^{x-2}} = \dfrac{3^{2x+2}}{3^{x-2}} = 3^{(2x+2)-(x-2)} = 3^{x+4}$ M1 (same base, division law)

5. $= 3^x \times 3^4 = 81u$ , as required A1*

Answer: (a) $9u$ (b) $\dfrac{u^3}{27}$ (c) $81u$ .

Test any index rewrite with a simple value: at $x = 0$ , $u = 1$ and $\dfrac{9}{3^{-2}} = 9 \times 9 = 81$ , which agrees with $81u$ .

In (c) the index subtraction $(2x + 2) - (x - 2)$ needs its bracket: written as $2x + 2 - x - 2$ , it loses the $+2$ and the answer comes out as $u$ .

Worked example 4: a model with two constants (exam length)

The height of a young tree, $H$ metres, $t$ years after it was planted is modelled by $H = a + bt^{\frac{2}{3}}$ , where $a$ and $b$ are constants.

One year after planting the height was 0.9 m, and eight years after planting it was 2.1 m.

(a) Find the value of $a$ and the value of $b$ . (3)

Making your method clear,

(b) find the time after planting at which the model predicts a height of 3.3 m, giving your answer to one decimal place. (2)

1. (a) $t = 1$ : $a + b = 0.9$ ; $t = 8$ : $8^{\frac{2}{3}} = 4$ , so $a + 4b = 2.1$ M1 (two equations, $8^{\frac{2}{3}}$ evaluated)

2. Subtract: $3b = 1.2$ M1 (solve the pair)

3. $b = 0.4$ , $a = 0.5$ A1

4. (b) $3.3 = 0.5 + 0.4t^{\frac{2}{3}} \Rightarrow t^{\frac{2}{3}} = 7 \Rightarrow t = 7^{\frac{3}{2}}$ M1 (isolate the power, raise to the reciprocal index)

5. $t = 18.52\ldots = 18.5$ years A1

Answer: $a = 0.5$ , $b = 0.4$ ; $t = 18.5$ years.

The constant $a$ is the model's height at planting ( $t = 0$ ): 0.5 m.

Raising both sides to the power $\frac{3}{2}$ undoes the power $\frac{2}{3}$ , because $\frac{2}{3} \times \frac{3}{2} = 1$ ; "Making your method clear" means this step must be on the page.

Exam tip

When a question defines a letter such as $u = 3^x$ , rewrite every base as a power of 3 first ( $9 = 3^2$ , $27 = 3^3$ , $\frac{1}{3} = 3^{-1}$ ), then split the index into " $x$ part" and "number part".

Give the answer with its number in front, such as $81u$ or $\frac{u^3}{27}$ , and test it with $x = 0$ before moving on.

Common mistakes

Adding to the index instead of multiplying. $3^{x+2}$ is $9u$ , not $u + 2$ : an added index is a multiplied factor. The wrong answer is tempting because "+2" appears in the question, so check with $x = 0$ : $3^2 = 9$ , not 2.

Merging the coefficient into the power. $16^{x+3} = 16^3 \times 16^x = 4096\,(2^x)^4$ , so with $m = 2^x$ it is $4096m^4$ , never $4096m$ . Keep the number and the power of the letter as separate factors and simplify each one.

Combining different bases. $4^x \times 2^x$ is not $8^{2x}$ : the laws only add indices of the same base. Write $4^x = 2^{2x}$ first, giving $2^{3x}$ .

Applying a power to only part of a bracket. $(3x^2)^3 = 27x^6$ ; the 3 is cubed too.

Check your understanding

1. Find the exact value of
(a) $32^{\frac{3}{5}}$ (1)
(b) $\left(\dfrac{9}{4}\right)^{-\frac{3}{2}}$ (2) Total [3]

2. Express each of the following in the form $kx^n$ , where $k$ and $n$ are constants in simplest form.
(a) $\dfrac{6x^2}{\sqrt[3]{x}}$ (1)
(b) $\left(2x^{-1}\right)^{-3}$ (2) Total [3]

3. Given that $w = 5^x$ ,
(a) express $5^{x-1}$ in terms of $w$ . (1)
(b) Express $125^{x+1}$ in terms of $w$ . (2)
(c) Show that $25^x + 5^{x+1} = w(w + 5)$ . (2) Total [5]

4. Show that $\dfrac{9^{n+1} \times 3^n}{27^n} = 9$ for every value of $n$ . (3) Total [3]

5. The number of visitors to a new website, $V$ thousand, $t$ weeks after launch, is modelled by $V = p + qt^{\frac{3}{2}}$ , where $p$ and $q$ are constants. After 1 week there were 2.6 thousand visitors and after 4 weeks there were 7.5 thousand.
(a) Find the value of $p$ and the value of $q$ . (3) Making your method clear,
(b) find the value of $t$ at which the model predicts 30 thousand visitors, giving your answer to one decimal place. (2) Total [5]

6. A student writes $2^3 \times 3^2 = 6^5$ . State whether the student is correct, giving a reason. (2) Total [2]

7. Show that, for $a \neq 0$ , the law $a^m \div a^n = a^{m-n}$ forces $a^{-2} = \dfrac{1}{a^2}$ . (2) Total [2]

Answers

1. (a) $\left(\sqrt[5]{32}\right)^3 = 2^3 = 8$ B1
(b) $\left(\dfrac{4}{9}\right)^{\frac{3}{2}}$ M1 (reciprocal for the negative index); $= \left(\dfrac{2}{3}\right)^3 = \dfrac{8}{27}$ A1

2. (a) $6x^{2 - \frac{1}{3}} = 6x^{\frac{5}{3}}$ B1
(b) $2^{-3}x^{3}$ M1; $= \dfrac{1}{8}x^3$ A1

3. (a) $\dfrac{w}{5}$ B1
(b) $125^{x+1} = 5^{3x+3} = 5^{3x} \times 5^3$ M1; $= 125w^3$ A1
(c) $25^x = 5^{2x} = w^2$ and $5^{x+1} = 5w$ M1; $w^2 + 5w = w(w + 5)$ A1*

4. $9^{n+1} = 3^{2n+2}$ , $27^n = 3^{3n}$ M1; $\dfrac{3^{2n+2} \times 3^n}{3^{3n}} = 3^{(2n + 2 + n) - 3n}$ M1; $= 3^2 = 9$ , which does not depend on $n$ A1*

5. (a) $p + q = 2.6$ ; $4^{\frac{3}{2}} = 8$ , so $p + 8q = 7.5$ M1; $7q = 4.9$ M1; $q = 0.7$ , $p = 1.9$ A1
(b) $30 = 1.9 + 0.7t^{\frac{3}{2}} \Rightarrow t^{\frac{3}{2}} = 40.14\ldots \Rightarrow t = (40.14\ldots)^{\frac{2}{3}}$ M1; $t = 11.7$ weeks A1

6. Not correct B1: the laws of indices only combine powers of the same base; $2^3 \times 3^2 = 8 \times 9 = 72$ , while $6^5 = 7776$ B1

7. $a^0 \div a^2 = a^{0-2} = a^{-2}$ M1; and $a^0 = 1$ , so $a^0 \div a^2 = \dfrac{1}{a^2}$ , hence $a^{-2} = \dfrac{1}{a^2}$ A1*

Laws of indices for all rational exponents

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