Indices

Unit 9 · Compulsory Mathematics · 8 min read

Indices

What the syllabus asks for 4
  • laws of indices
  • simplification using the laws
  • negative and zero indices
  • exponential equations

An index is shorthand for repeated multiplication: 2^5 means 2\times2\times2\times2\times2. The shorthand is not just a saving of ink — it turns multiplication into addition of indices, division into subtraction, and powers of powers into multiplication, which is why indices appear in every later chapter of algebra, in scientific notation, and in the arithmetic of growth. This chapter sets out the laws, shows how to simplify expressions with them, and shows how to solve the small exponential equations the syllabus asks for.

What this chapter asks of you

You should be able to state and use the five laws of indices, evaluate negative and zero indices, simplify an algebraic expression involving indices, and solve a simple exponential equation.

9.1 The laws of indices

Law Statement Example
Product a^m\times a^n=a^{m+n} 2^3\times2^4=2^7=128
Quotient a^m\div a^n=a^{m-n} 5^6\div5^4=5^2=25
Power of a power (a^m)^n=a^{mn} (3^2)^3=3^6=729
Power of a product (ab)^n=a^nb^n (2x)^3=8x^3
Power of a quotient \left(\frac ab\right)^n=\frac{a^n}{b^n} \left(\frac23\right)^2=\frac49

In a^n, a is the base and n is the index (power). Each law comes from writing the powers out in full:

x^2\times x^3=(x\times x)\times(x\times x\times x)=x^5=x^{2+3}

x^5\div x^2=\frac{x\times x\times x\times x\times x}{x\times x}=x^3=x^{5-2}

(x^2)^3=x^2\times x^2\times x^2=x^{2+2+2}=x^{2\times3}=x^6

Two more results follow from the quotient law and are used constantly:

a^0=1\qquad(a\ne0),\qquad a^{-n}=\frac{1}{a^n}

The first comes from a^m\div a^m=a^{m-m}=a^0, and the quotient is 1. The second comes from a^0\div a^n=a^{0-n}=a^{-n}, and dividing 1 by a^n gives \frac{1}{a^n}.

a^0=1 is not 0

Any non-zero number raised to the power zero is 1. In particular 7^0=1, x^0=1 and (3xy)^0=1. But 0^0 is not defined, and 0^5 is 0 — the zero in the index and the zero in the base are different things.

Using the product and quotient laws

Problem. Simplify \frac{3^5\times3^2}{3^4}.

Solution. Add the indices in the numerator first:

3^5\times3^2=3^{5+2}=3^7

Then subtract: 3^7\div3^4=3^{7-4}=3^3=27.

Answer. 27.

Check. 3^5=243 and 3^2=9, so the numerator is 2187; and 2187\div81=27 ✓.

An algebraic simplification

Problem. Simplify \frac{12x^5y^3}{4x^2y}.

Solution. Divide the numbers, then use the quotient law on each letter:

\frac{12}{4}=3,\qquad \frac{x^5}{x^2}=x^{5-2}=x^3,\qquad \frac{y^3}{y}=y^{3-1}=y^2

\frac{12x^5y^3}{4x^2y}=3x^3y^2

Answer. 3x^3y^2.

Check. Put x=2, y=3: the left side is \frac{12\times32\times27}{4\times4\times3}=\frac{10368}{48}=216, and the right side is 3\times8\times9=216 ✓.

Zero and negative indices

Problem. Evaluate 5^0+2^{-3} and simplify \left(\frac{2x}{y}\right)^{-2}.

Solution. 5^0=1 and 2^{-3}=\frac{1}{2^3}=\frac18, so

5^0+2^{-3}=1+\frac18=1\frac18=\frac98

For the second, a negative index on a fraction turns it upside down:

\left(\frac{2x}{y}\right)^{-2}=\left(\frac{y}{2x}\right)^{2} =\frac{y^2}{4x^2}

Answer. 1\frac18 and \frac{y^2}{4x^2}.

Check. For the second, put x=1 and y=2. The left side is \left(\frac{2}{2}\right)^{-2}=1^{-2}=1, and the right side is \frac{4}{4\times1}=1. They agree ✓. Using x=2, y=1 instead: the left side is 4^{-2}=\frac{1}{16}, and the right side is \frac{1}{4\times4}=\frac{1}{16} ✓.

9.2 Simplifying expressions with several laws

The order to work in: deal with the brackets first, then the powers of powers, then the products and quotients.

A three-step simplification

Problem. Simplify \left(\frac{2a^3b^{-2}}{4ab}\right)^2.

Solution. Step 1 — simplify inside the bracket. Divide the numbers: \frac{2}{4}=\frac12. For a: a^{3-1}=a^2. For b: b^{-2-1}=b^{-3}. So inside we have \frac{a^2b^{-3}}{2}, that is \frac{a^2}{2b^3}.

Step 2 — square everything.

\left(\frac{a^2}{2b^3}\right)^2=\frac{a^{4}}{4b^{6}}

Answer. \frac{a^4}{4b^6}.

Check. Put a=1, b=1: the original is \left(\frac{2}{4}\right)^2=\frac14, and the answer is \frac14 ✓.

Powers of a product

Problem. Simplify (2x^2y)^3\times(3xy^2)^2.

Solution. Apply the power to each factor:

(2x^2y)^3=2^3x^6y^3=8x^6y^3,\qquad (3xy^2)^2=9x^2y^4

Now multiply, adding indices of like bases:

8x^6y^3\times9x^2y^4=72x^{6+2}y^{3+4}=72x^8y^7

Answer. 72x^8y^7.

Check. Put x=y=1: the original is 2^3\times3^2=8\times9=72, and the answer gives 72 ✓.

9.2.1 Letters in the index

The laws work in exactly the same way when the indices are algebraic expressions. Put brackets round each index before subtracting.

Algebraic indices

Problem. Simplify (a) x^{3n-2}\div x^{2n-5} (b) x^{a-b}\times x^{b-c}\times x^{c-a} (c) \frac{x^{4n-2}}{x^{2(2n-1)}}.

Solution. (a) x^{(3n-2)-(2n-5)}=x^{3n-2-2n+5}=x^{n+3}.

(b) x^{a-b+b-c+c-a}=x^0=1.

(c) x^{(4n-2)-(4n-2)}=x^0=1.

Bracket the index you subtract

x^{3n-2}\div x^{2n-5} is x^{(3n-2)-(2n-5)}. Writing x^{3n-2-2n-5}=x^{n-7} forgets that the minus sign changes both terms of 2n-5.

Proving an identity

Problem. Prove that \frac{x^{p-q+1}\times x^{q-r+1}\times x^{r-p+1}}{x^3}=1.

Solution. LHS =\frac{x^{(p-q+1)+(q-r+1)+(r-p+1)}}{x^3}=\frac{x^{3}}{x^3}=x^{3-3}=x^0=1= RHS. Proved.

9.3 Exponential equations

An exponential equation has the unknown in the index, such as 2^x=32. The method is to write both sides as powers of the same base and then equate the indices.

a^m=a^n\;\Longrightarrow\;m=n\qquad(a>0,\ a\ne1)

Solving by making the bases equal

Problem. Solve 3^x=81.

Solution. Write 81 as a power of 3: 81=3^4. So

3^x=3^4\;\Longrightarrow\;x=4

Answer. x=4.

Check. 3^4=81 ✓.

An equation with a second term

Problem. Solve 2^{x+1}=64.

Solution. 64=2^6, so 2^{x+1}=2^6 and therefore x+1=6, giving x=5.

Answer. x=5.

Check. 2^{5+1}=2^6=64 ✓.

An equation with a power on both sides

Problem. Solve 9^{x}=27^{2}.

Solution. Write both numbers as powers of 3: 9=3^2 and 27=3^3. So

(3^2)^x=(3^3)^2\;\Longrightarrow\;3^{2x}=3^{6}

Equating indices: 2x=6, so x=3.

Answer. x=3.

Check. 9^3=729 and 27^2=729 ✓.

The base must be the same before you compare indices

From 2^x=8^2 you may not write x=2. Change the right side to the base 2 first: 8^2=(2^3)^2=2^6, so x=6. Comparing indices of different bases is not allowed.

9.4 The laws of indices in one place

figure
Figure 1. How fast a power grows. Doubling just ten times multiplies a quantity by more than a thousand — this is why indices are used for very large and very small numbers.

Law Statement Example
Product a^m \times a^n = a^{m+n} 2^3 \times 2^4 = 2^7
Quotient \frac{a^m}{a^n} = a^{m-n} \frac{5^6}{5^2} = 5^4
Power of a power (a^m)^n = a^{mn} (3^2)^3 = 3^6
Power of a product (ab)^m = a^m b^m (2x)^3 = 8x^3
Power of a quotient \left(\frac{a}{b}\right)^m = \frac{a^m}{b^m} \left(\frac{2}{3}\right)^2 = \frac{4}{9}
Zero index a^0 = 1, a \neq 0 7^0 = 1
Negative index a^{-n} = \frac{1}{a^n} 4^{-2} = \frac{1}{16}
The three commonest slips

a^m + a^n is not a^{m+n} — the product law applies to multiplication, never to addition. (-2)^2 = 4 but -2^2 = -4, because the index binds tighter than the minus sign. And (a+b)^2 \neq a^2 + b^2.

Working with mixed indices

Question. Simplify \dfrac{(2^3)^2 \times 2^{-4}}{2^{-1}}.

Answer. Numerator: (2^3)^2 = 2^6, so 2^6 \times 2^{-4} = 2^{2}. Dividing: \frac{2^2}{2^{-1}} = 2^{2-(-1)} = 2^3 = \textbf{8}.

Solving for an index

Question. Solve 3^{2x-1} = 81.

Answer. Write both sides with the same base: 81 = 3^4. So 2x - 1 = 4 \Rightarrow 2x = 5 \Rightarrow x = \frac{5}{2} = \textbf{2.5}.

Activity — the folding paper

Fold a sheet of paper in half as many times as you can, recording the number of layers after each fold: 1, 2, 4, 8, \ldots Write the count as a power of 2. Discussion: Most people stop at six or seven folds. If you could fold it 30 times, the stack would be 2^{30} sheets thick — work out roughly how many kilometres that is, taking one sheet as 0.1 mm.

Chapter summary

  • a^m\times a^n=a^{m+n}; a^m\div a^n=a^{m-n}; (a^m)^n=a^{mn}; (ab)^n=a^nb^n; \left(\frac ab\right)^n=\frac{a^n}{b^n}.
  • a^0=1 for every non-zero a, and a^{-n}=\frac{1}{a^n}.
  • Simplify inside brackets first, then apply powers, then multiply or divide like bases by adding or subtracting indices.
  • An exponential equation is solved by writing both sides as powers of the same base and equating the indices.

Practice questions

Group A — Objective (1 mark each)

  1. 2^3\times2^4 equals [1]a2^7b2^{12}c4^7d2^1
  2. 5^0 equals [1]a0b1c5dnot defined
  3. 3^{-2} equals [1]a-9b-6c\frac19d\frac13
  4. (x^2)^5 equals [1]ax^7bx^{10}cx^{25}d2x^5
  5. If 2^x=32, then x is [1]a4b5c6d16
  6. \frac{a^6}{a^2} equals [1]aa^3ba^4ca^8da^{12}
Answers to Group A

1. (a) — the indices add.

2. (b) — any non-zero number to the power zero is 1.

3. (c) — 3^{-2}=\frac{1}{9}.

4. (b) — multiply the indices: 2\times5=10.

5. (b) — 32=2^5.

6. (b) — subtract: 6-2=4.

Group B — Short answer (2 marks each)

  1. Simplify 4^3\times4^2\div4^4. [2]
  2. Simplify \frac{15a^4b^3}{5a^2b}. [2]
  3. Evaluate \left(\frac{3}{4}\right)^{-2}. [2]
  4. Solve 5^x=125. [2]
Answers to Group B

1. 4^{3+2-4}=4^1=4.

2. 3a^{2}b^{2}.

3. \left(\frac{4}{3}\right)^2=\frac{16}{9}=1\frac79.

4. 125=5^3, so x=3.

Group C — Long answer (4 marks each)

  1. Simplify \left(\frac{3x^2y^{-1}}{6xy^2}\right)^{-2} and give the answer with positive indices only. [4]
  2. Solve 4^{x+1}=64 and 9^{x}=\frac{1}{27}\times3^{x+1}. [4]
Answers to Group C

1. Step 1 — simplify inside. \frac{3}{6}=\frac12; for x: x^{2-1}=x; for y: y^{-1-2}=y^{-3}. So inside the bracket we have \frac{x}{2y^3}, that is \frac{xy^{-3}}{2}. Step 2 — apply the power -2. \left(\frac{x}{2y^3}\right)^{-2}=\left(\frac{2y^3}{x}\right)^{2} =\frac{4y^6}{x^2}. Check. Put x=y=1: the original is \left(\frac{3}{6}\right)^{-2}=\left(\frac12\right)^{-2}=4, and the answer gives \frac41=4 ✓.

2. (i) 64=4^3, so 4^{x+1}=4^3 and x+1=3, giving x=2. Check: 4^3=64 ✓. (ii) Write everything as powers of 3: 9^x=3^{2x} and \frac{1}{27}=3^{-3}. So 3^{2x}=3^{-3}\times3^{x+1}=3^{x-2}. Equating indices: 2x=x-2, so x=-2. Check: 9^{-2}=\frac{1}{81} and \frac{1}{27}\times3^{-1}=\frac{1}{81} ✓.

Extra practice

Group B (2 marks)

11. Simplify: \dfrac{x^5 \times x^{-2}}{x^{-3}}.

12. If a=1, b=2, c=3, find the value of a^b\times b^c\times c^a.

Group C (4 marks)

13. Solve: (i) 2^{x+2} = 128 (ii) 5^{2x} = \dfrac{1}{125}.

Extra practice — answers

11. x^{5-2} = x^3; then \frac{x^3}{x^{-3}} = x^{3+3} = x^6.

12. 1^2\times2^3\times3^1=1\times8\times3=24.

13. (i) 128 = 2^7, so x + 2 = 7 and x = 5. (ii) \frac{1}{125} = 5^{-3}, so 2x = -3 and x = -\frac{3}{2} = -1.5.

Objective practice

Choose the correct answer. [5×1=5]

1. 3^4 \times 3^2 equals (a) 3^6 (b) 3^8 (c) 9^6 (d) 3^2

2. 5^0 + 2^{-1} equals (a) 0 (b) 1 (c) 1.5 (d) 2

3. \frac{a^{n-2}}{a^{n+2}} equals (a) a^4 (b) \frac{1}{a^4} (c) a^{2n} (d) 1

4. If 3^x = 81, then x is (a) 3 (b) 4 (c) 9 (d) 27

5. (-3)^2 equals (a) -9 (b) 9 (c) -6 (d) 6

Objective practice — answers

1. (a) — add the indices.

2. (c) — 1+0.5=1.5.

3. (b) — a^{(n-2)-(n+2)}=a^{-4}=\frac{1}{a^4}.

4. (b) — 3^4=81.

5. (b) — the bracket squares the sign too.

Practise more: Mathematics question bank