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How do the laws of indices work, and how do we write very large or small numbers in standard form?

Apply the laws of indices including zero, negative and fractional powers, and express and calculate with numbers in standard form

A focused answer to the O-Level E-Maths outcome on indices and standard form. The laws of indices, zero, negative and fractional powers, and writing and calculating with numbers in standard form.

Generated by Claude Opus 4.88 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

SEAB wants you to apply the laws of indices, including zero, negative and fractional powers, and to write and compute with numbers in standard form (scientific notation). These tools let you handle powers compactly and deal with very large and very small quantities without long strings of zeros.

The answer

The laws of indices

For the same base, the index laws are:

am×an=am+n,aman=amn,(am)n=amna^m \times a^n = a^{m+n}, \qquad \frac{a^m}{a^n} = a^{m-n}, \qquad (a^m)^n = a^{mn}

Add indices when multiplying, subtract when dividing, and multiply when raising a power to a power. A power of a product distributes: (ab)n=anbn(ab)^n = a^n b^n.

Zero and negative indices

Any non-zero number to the power zero is 11, so 70=17^0 = 1. A negative index is a reciprocal:

an=1ana^{-n} = \frac{1}{a^n}

so 23=182^{-3} = \dfrac{1}{8}.

Fractional indices

The denominator of a fractional index is a root and the numerator is a power:

a1/n=an,am/n=(an)ma^{1/n} = \sqrt[n]{a}, \qquad a^{m/n} = \left(\sqrt[n]{a}\right)^{m}

For example 82/3=(83)2=22=48^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4.

Standard form

Standard form writes a number as A×10nA \times 10^n, where 1A<101 \le A < 10 and nn is an integer. Large numbers have positive nn; small numbers have negative nn. So 4500000=4.5×1064\,500\,000 = 4.5 \times 10^6 and 0.00072=7.2×1040.00072 = 7.2 \times 10^{-4}.

Calculating in standard form

Multiply or divide the leading numbers and add or subtract the powers of ten, then adjust so the leading number lies between 11 and 1010:

(6×104)×(3×105)=18×109=1.8×1010(6 \times 10^4) \times (3 \times 10^5) = 18 \times 10^9 = 1.8 \times 10^{10}

Adding and subtracting in standard form

Multiplication and division in standard form are straightforward, but addition and subtraction need the powers of ten to match first. Rewrite the numbers so they share the same power of ten, then add or subtract the leading parts, and finally adjust back to proper standard form. To compute 3×104+5×1033 \times 10^4 + 5 \times 10^3, rewrite the second as 0.5×1040.5 \times 10^4, so the sum is 3.5×1043.5 \times 10^4. The step students skip is aligning the powers, which is essential because 10410^4 and 10310^3 are different units of size that cannot be combined directly.

Comparing numbers in standard form

Standard form makes comparing very large or very small numbers quick: compare the powers of ten first, and only if those are equal compare the leading numbers. So 4×1064 \times 10^6 is larger than 9×1059 \times 10^5 despite the smaller leading digit, because 106>10510^6 > 10^5. For negative powers (small numbers), a less negative power is larger, so 2×1032 \times 10^{-3} exceeds 8×1058 \times 10^{-5}. Ordering a list of numbers by their power of ten first, then by leading digit, is the reliable method and a frequent E-Maths task that catches out anyone who only looks at the leading number.

Examples in context

Example 1. Distances in astronomy. The distance from the Earth to the Sun is about 1.5×1011 m1.5 \times 10^{11}\ \text{m}. Standard form lets scientists write and compare such huge distances without dozens of zeros, and arithmetic on them reduces to handling the leading numbers and the powers of ten.

Example 2. Sizes in biology. A bacterium might be 2×106 m2 \times 10^{-6}\ \text{m} across. Negative powers of ten make tiny measurements manageable, and dividing two such measurements quickly gives a ratio of sizes.

Try this

Q1. Evaluate 50+325^{0} + 3^{-2}. [2 marks]

  • Cue. 50=15^0 = 1 and 32=193^{-2} = \dfrac{1}{9}, so the sum is 1191\dfrac{1}{9} or 109\dfrac{10}{9}.

Q2. Write 0.0000360.000\,036 in standard form. [1 mark]

  • Cue. Move the point so the leading digit is between 11 and 1010: 3.6×1053.6 \times 10^{-5}.

Q3. Evaluate 163/416^{3/4}. [2 marks]

  • Cue. Fourth root then cube: (164)3=23=8\left(\sqrt[4]{16}\right)^3 = 2^3 = 8.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original3 marksSimplify a5×a2a4\dfrac{a^5 \times a^{-2}}{a^{4}}, giving your answer with a positive index.
Show worked answer →

Add indices when multiplying: a5×a2=a5+(2)=a3a^5 \times a^{-2} = a^{5 + (-2)} = a^{3}.

Subtract indices when dividing: a3a4=a34=a1\dfrac{a^3}{a^4} = a^{3 - 4} = a^{-1}.

With a positive index this is 1a\dfrac{1}{a}.

Markers reward adding indices for the product, subtracting for the quotient, and rewriting the negative index as a fraction.

Original4 marksThe mass of one molecule is 3.0×1023 g3.0 \times 10^{-23}\ \text{g}. (a) Find the mass of 5.0×1085.0 \times 10^{8} molecules, giving your answer in standard form. (b) Express this mass in grams as an ordinary number.
Show worked answer →

(a) Multiply the numbers and add the powers of ten: (3.0×5.0)×1023+8=15×1015(3.0 \times 5.0) \times 10^{-23 + 8} = 15 \times 10^{-15}.

In standard form the number must be between 11 and 1010, so 15×1015=1.5×1014 g15 \times 10^{-15} = 1.5 \times 10^{-14}\ \text{g}.

(b) As an ordinary number, 1.5×1014=0.000000000000015 g1.5 \times 10^{-14} = 0.000\,000\,000\,000\,015\ \text{g}.

Markers reward multiplying the leading numbers, adding the indices, correcting to proper standard form, and the ordinary-number form.

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