This note covers the laws of indices (including zero, negative and fractional powers), standard form, and surds. Indices and standard form are used for very large and very small quantities in science and computing, and surds let you keep answers exact. Surd work (simplifying, and rationalising a denominator) is Extended only.
Key ideas
An index (power) tells you how many times a base is multiplied by itself. The same few laws work for every base, as long as the bases match.
- Multiplying powers of the same base: add the indices. Dividing: subtract them. A power of a power: multiply the indices.
- Any non-zero number to the power 0 equals 1.
- A negative index means "reciprocal", not "negative number". For example \(2^{-3}=\dfrac{1}{8}\).
- A fractional index means a root and a power. The denominator is the root and the numerator is the power: \(a^{m/n}=\left(\sqrt[n]{a}\right)^m\). It is usually easiest to take the root first, while the numbers are small.
- To solve an equation such as \(3^x=\dfrac{1}{81}\), write both sides as powers of the same base, then compare the indices.
Standard form. A number is in standard form when it is written as \(a\times10^n\) with \(1\le a\lt 10\) and \(n\) an integer. Large numbers have positive \(n\) and numbers smaller than 1 have negative \(n\). To multiply or divide, handle the numbers \(a\) and the powers of 10 separately, then tidy up so that \(a\) is between 1 and 10 again. To add or subtract, first write both numbers with the same power of 10.
Surds (Extended). A surd is a root that cannot be written as a fraction, such as \(\sqrt{2}\) or \(\sqrt{7}\). Surds are exact; a decimal from the calculator is only an approximation.
- To simplify, pull out the largest square factor: \(\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}\).
- You can only add or subtract "like" surds, such as \(4\sqrt{3}+\sqrt{3}=5\sqrt{3}\). Simplify each surd first, as it may then match another.
- To rationalise \(\dfrac{a}{\sqrt{b}}\), multiply top and bottom by \(\sqrt{b}\).
- To rationalise \(\dfrac{a}{b+\sqrt{c}}\), multiply top and bottom by the conjugate \(b-\sqrt{c}\). The denominator becomes \(b^2-c\), which is rational.
Using your calculator. Most scientific calculators have a key for entering "times ten to the power" (often marked EXP or \(\times10^x\)). Use it to type a number such as \(4.5\times10^{6}\) in one go, rather than typing \(\times10\) and a power separately. This avoids bracket mistakes in a longer calculation. When you copy the answer down, write it out fully in standard form rather than copying the calculator display.
Choosing the right form. Write a number in standard form when the question asks for it or when it is extremely large or small. Use a surd when the question asks for an exact answer. Use a decimal only when the question asks for a rounded value, and round only at the very end.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(a^m\times a^n=a^{m+n}\) | Same base, multiplying: add the indices. |
| \(a^m\div a^n=a^{m-n}\) | Same base, dividing: subtract the indices. |
| \((a^m)^n=a^{mn}\) | Power of a power: multiply. Applies to every factor inside the bracket, including numbers. |
| \(a^0=1\) | True for every \(a\ne0\). |
| \(a^{-n}=\dfrac{1}{a^n}\) | Negative index: take the reciprocal. |
| \(\left(\dfrac{a}{b}\right)^{-n}=\left(\dfrac{b}{a}\right)^n\) | Negative index on a fraction: flip the fraction. |
| \(a^{m/n}=\left(\sqrt[n]{a}\right)^m\) | Fractional index: root first, then power. |
| \(a\times10^n\), \(1\le a\lt10\) | Standard form for very large or very small numbers. |
| \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\) | Combining or splitting surds (Extended). |
| \((b+\sqrt{c})(b-\sqrt{c})=b^2-c\) | Used to rationalise a denominator of the form \(b+\sqrt{c}\) (Extended). |
Worked examples
Example 1 Evaluate \(\left(\dfrac{8}{27}\right)^{-2/3}\) without a calculator.
The negative index means flip the fraction: \(\left(\dfrac{27}{8}\right)^{2/3}\).
The denominator 3 of the index is the root, so take the cube root of top and bottom first: \(\sqrt[3]{27}=3\) and \(\sqrt[3]{8}=2\), giving \(\left(\dfrac{3}{2}\right)^{2}\).
The numerator 2 is the power: \(\dfrac{3^2}{2^2}=\dfrac{9}{4}\).
Answer: \(\dfrac{9}{4}\) (or 2.25)
Example 2 A hard drive stores \(4.5\times10^{6}\) files. Each file is \(3.2\times10^{-3}\) gigabytes. (a) Find the total storage used, in standard form. (b) A different file of \(6.4\times10^{-2}\) gigabytes is split into \(1.6\times10^{5}\) equal pieces. Find the size of one piece. (c) Work out \(3.2\times10^{5}+4.5\times10^{4}\).
(a) Multiply the numbers and the powers of 10 separately: \((4.5\times3.2)\times10^{6+(-3)}=14.4\times10^{3}\).
14.4 is not between 1 and 10, so write \(14.4=1.44\times10^{1}\). Then \(1.44\times10^{1}\times10^{3}=1.44\times10^{4}\) gigabytes.
(b) Divide the numbers: \(6.4\div1.6=4\). Subtract the indices: \(-2-5=-7\). One piece is \(4\times10^{-7}\) gigabytes.
(c) Write both with the same power of 10: \(4.5\times10^{4}=0.45\times10^{5}\). Add: \((3.2+0.45)\times10^{5}=3.65\times10^{5}\).
Answer: (a) \(1.44\times10^{4}\) GB; (b) \(4\times10^{-7}\) GB; (c) \(3.65\times10^{5}\)
Example 3 (Extended) (a) Simplify \(\sqrt{72}-\sqrt{18}\). (b) Rationalise the denominator of \(\dfrac{10}{3-\sqrt{5}}\) and write the answer in the form \(\dfrac{a+b\sqrt{5}}{c}\).
(a) \(\sqrt{72}=\sqrt{36\times2}=6\sqrt{2}\) and \(\sqrt{18}=\sqrt{9\times2}=3\sqrt{2}\).
Now the surds are alike, so \(6\sqrt{2}-3\sqrt{2}=3\sqrt{2}\).
(b) The conjugate of \(3-\sqrt{5}\) is \(3+\sqrt{5}\). Multiply top and bottom by it:
\[\frac{10}{3-\sqrt{5}}\times\frac{3+\sqrt{5}}{3+\sqrt{5}}=\frac{10(3+\sqrt{5})}{9-5}=\frac{10(3+\sqrt{5})}{4}\]
Cancel a factor of 2: \(\dfrac{5(3+\sqrt{5})}{2}=\dfrac{15+5\sqrt{5}}{2}\).
Check with decimals: \(10\div(3-2.236)\approx13.09\) and \((15+11.18)\div2\approx13.09\).
Answer: (a) \(3\sqrt{2}\); (b) \(\dfrac{15+5\sqrt{5}}{2}\)
Common errors
- Writing \(2^{-3}=-8\) or \(-6\). A negative index gives a reciprocal, so \(2^{-3}=\frac18\).
- Leaving an answer such as \(14.4\times10^3\) as it is. Standard form needs the first number between 1 and 10.
- Applying a power to only part of a bracket: \((2x^3)^4=16x^{12}\), not \(2x^{12}\).
- Writing \(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\). This is false: \(\sqrt{9+16}=5\), but \(3+4=7\).
- In rationalising, multiplying by the same sign (\(3-\sqrt5\)) instead of the conjugate (\(3+\sqrt5\)). The root then stays in the denominator.
Exam tip: With a fractional or negative index, deal with the sign first (flip), then the root, then the power. Show each step so that a slip in the last step does not cost the earlier ones.
Check yourself
- Simplify \(\dfrac{(2x^3)^4}{8x^5}\).
Answer
\((2x^3)^4=16x^{12}\). Then \(16x^{12}\div8x^5=2x^{12-5}=2x^7\). - Write 0.00072 in standard form, and write \(5.8\times10^{7}\) as an ordinary number.
Answer
Move the decimal point 4 places to get 7.2, so \(0.00072=7.2\times10^{-4}\). Also \(5.8\times10^{7}=58\,000\,000\). - Solve \(3^x=\dfrac{1}{81}\).
Answer
\(81=3^4\), so \(\frac{1}{81}=3^{-4}\). Comparing indices, \(x=-4\). - (Extended) Express \((2+\sqrt{3})^2\) in the form \(a+b\sqrt{3}\).
Answer
\((2+\sqrt3)(2+\sqrt3)=4+2\sqrt3+2\sqrt3+3=7+4\sqrt{3}\).