SPI Mathematics · Surds, Indices and Simplification

Laws of Indices

Powers of the same base multiply by adding their indices, divide by subtracting them, and a power of a power multiplies them.

All Surds, Indices and Simplification notes

Why this matters

Nine questions in the bank, seven EASY and two MODERATE, and one is from the 2022 paper. Every one is settled by a few laws: add the indices to multiply, multiply them for a power of a power, and know what a zero or a negative index means.

Concept 1 of 2: Multiplying, dividing and raising powers

a^m

is just

a

written down

m

times and multiplied. Put

m

of them next to

n

more and you have

m + n

. Take a power of a power and you copy the whole block again and again, so the indices multiply.

Definition

$a^m \times a^n = a^{m + n}$ : same base, add the indices.
$a^m \div a^n = a^{m - n}$ : same base, subtract the indices.
$(a^m)^n = a^{mn}$ : a power of a power, multiply the indices.
$(ab)^n = a^n b^n$ and $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$ : the same index on different bases.
The first three laws need the same base. $2^3 \times 3^2$ cannot be joined into one power.

Laws of indices

a^m \times a^n = a^{m+n} \qquad a^m \div a^n = a^{m-n} \qquad (a^m)^n = a^{mn} \qquad (ab)^n = a^n b^n

Worked example

Simplify

\dfrac{2^7 \times 2^3}{(2^2)^4}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SPI · 2022 · SPI Entrance Examination 2022 · Q67Moderate

Example 1 · Surds, Indices and Simplification · Laws of Indices

\left[\left(x^{y+1}\right)^{\frac{y^2}{y^2-1}}\right]^{1-\frac{1}{y}} = ?

Adding indices for a power of a power

(a^2)^3

a^6

, not

a^5

. Add the indices only when two powers of the same base are multiplied.

Joining different bases

2^3 \times 3^2 = 8 \times 9 = 72

, not

6^5

. The laws that add or subtract indices work only for one base.

Spreading a power over a sum

(a + b)^2

is not

a^2 + b^2

(2 + 3)^2 = 25

, but

4 + 9 = 13

. A power spreads over a product, never over a sum.

Drill 4 more on multiplying, dividing and raising powers

Concept 2 of 2: Zero and negative indices

Each step down in the index divides by the base once:

2^3 = 8

2^2 = 4

2^1 = 2

, so

2^0 = 1

and

2^{-1} = \frac{1}{2}

. A negative index means 'one over', not 'minus'.

Definition

$a^0 = 1$ for any non-zero $a$ .
$a^{-n} = \frac{1}{a^n}$ , and $\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$ : turn the fraction over and make the index positive.
A negative base to an even power is positive; to an odd power it is negative: $(-2)^4 = 16$ , $(-2)^3 = -8$ .
Brackets matter: $-2^4 = -16$ , because only the 2 is raised to the power.

Zero and negative indices

a^0 = 1 \ (a \ne 0) \qquad a^{-n} = \frac{1}{a^n} \qquad \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^{n}

Worked example

Find the value of

\left(\dfrac{2}{5}\right)^{-2} + 7^0

Practice this conceptself-check · 5 quick reps

The same idea in a practice question:

Practice questionEasy

Example 2 · Surds, Indices and Simplification · Laws of Indices

If x is a whole number, then

8^x

ends with an even digit, except for which value of x?

Reading a zero index as 0

a^0

is 1, not 0, for every non-zero a. Only

0^0

is left undefined.

Reading a negative index as a negative number

2^{-3} = \frac{1}{8}

, not

-8

. The minus sign in the index turns the number into its reciprocal.

Losing the sign of a negative base

\left(-\frac{3}{4}\right)^2 = \frac{9}{16}

is positive, but

\left(-\frac{3}{4}\right)^3 = -\frac{27}{64}

is negative. Check whether the power is even or odd.

Drill 3 more on zero and negative indices

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Multiplying, dividing and raising powers
Laws of indices
$a^m \times a^n = a^{m+n} \qquad a^m \div a^n = a^{m-n} \qquad (a^m)^n = a^{mn} \qquad (ab)^n = a^n b^n$
Zero and negative indices
Zero and negative indices
$a^0 = 1 \ (a \ne 0) \qquad a^{-n} = \frac{1}{a^n} \qquad \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^{n}$

Watch out for (6)

Adding indices for a power of a power
→ Multiplying, dividing and raising powers
Joining different bases
→ Multiplying, dividing and raising powers
Spreading a power over a sum
→ Multiplying, dividing and raising powers
Reading a zero index as 0
→ Zero and negative indices
Reading a negative index as a negative number
→ Zero and negative indices
Losing the sign of a negative base
→ Zero and negative indices

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