9 Indices
Indices is another word for powers. In this section we move beyond the idea that powers are just repeated multiplications.
9.1 Index notation
Being comfortable moving between different ways to write powers helps when rearranging algebra.
- \(x^0 = 1\) except when \(x=0\) then it’s undefined
- \(x^{-n} = \frac{1}{x^n}\)
- \(x^{\frac{1}{n}} = \sqrt[n]x\)
Here are some examples:
\[ 2^{-3} = \frac{1}{2^3} = \frac{1}{8} \] More generally. \[ x^{-3} = \frac{1}{x^3} \] Anything to the power of zero is \(1\): \[ \pi^0 = 1 \] Remember good old \(\pi\)? From working stuff out about circles \(\pi = 3.14159...\)
We can write square roots:
\[ 16^{\frac{1}{2}} = \sqrt{16} = \pm4 \]
When taking square roots remember there are two possible solutions. Since in the above example \(4 \times 4 = 16\) and \(-4 \times -4 = 16\). So either answer is just fine.
Here’s an example of a cube root.
\[ 8^{\frac{1}{3}} = \sqrt[3]{8} = 2 \]
9.1.1 Combinations, roots and powers
A roots and powers can be combined. If a number is raised to the power of a fraction you find the root corresponding to the denominator and then raise it to the power of the numerator. For example:
\[ 8^{\frac{2}{3}} = (\sqrt[3]{8})^2 = (2)^2 = 4 \]
Cube root, because of the \(3\) in the denominator, then square the answer because of the \(2\) in the numerator. This sequence could be done the other way around, square first then cube root, I choose this way since the numbers stay smaller.
9.1.2 Reciprocals
If you raise a number to the power of \(-1\) you find it’s reciprocal (you flip it). For example:
\[ \left(\frac{2}{3}\right)^{-1} = \frac{3}{2} \]
9.1.3 But why?
Just like we did with negative numbers we can extend the idea of what a power means by following a pattern. Here’s a pattern to justify \(x^0 = 1\) and \(x^{-n} = \frac{1}{x^n}\).
\[ \begin{aligned} 10^3 &= 10 \times 10 \times 10 & = 1000 \\ 10^2 &= 10 \times 10 & = 100 \\ 10^1 &= 10 & = 10 \\ 10^0 &= 1 & = 1 \\ 10^{-1} &= \frac{1}{10} & = 0.1 \\ 10^{-2} &= \frac{1}{10 \times 10} & = 0.01 \\ 10^{-3} &= \frac{1}{10 \times 10 \times 10} & = 0.001 \end{aligned} \]
I’ll come back to the justification about square roots after the next section.
9.2 Rules of indices
There is a neat set of rules we can use when combining numbers with indices:
- \(x^n \times x^m = x^{n+m}\)
- \(x^n \div x^m = x^{n-m}\)
- \((x^n)^m = x^{n \times m}\)
When you multiply terms you add the powers.
\[ \begin{aligned} 3x^4 \times 5x^6 &= 3 \times 5 \times x^4 \times x^5 \\ &= 15 \times x^{4+5} \\ &= 15x^9 \end{aligned} \]
Lets put it all together with a complicated example:
To rewrite \(\dfrac{\sqrt[4]{x^5x^3}}{\sqrt[3]{x} \sqrt[6]{x^3}}\) in the form \(x^n\), we need to use the following rules:
- \(a^n a^m = a^{n+m}\);
- \(\sqrt[n]{a} = a^{1/n}\);
- \(\left(a^n\right)^m = a^{n \times m}\);
- \(\frac{a^n}{a^m} = a^{n-m}\).
We will simplify the numerator and denominator separately to make the steps clearer. Firstly, applying rule 1, then rule 2, and then rule 3 to the numerator:
\[ \begin{aligned} \dfrac{\sqrt[4]{x^5x^3}}{\sqrt[3]{x} \sqrt[6]{x^3}} &= \dfrac{\sqrt[4]{x^8}}{\sqrt[3]{x} \sqrt[6]{x^3}} \\ &= \dfrac{(x^8)^{1/4}}{\sqrt[3]{x} \sqrt[6]{x^3}} \\ &= \dfrac{x^2}{\sqrt[3]{x} \sqrt[6]{x^3}} \end{aligned} \]
To simplify the denominator, we want to apply rule 2, then rule 3, and then rule 1:
\[ \begin{aligned} \dfrac{x^2}{\sqrt[3]{x} \sqrt[6]{x^3}} &= \dfrac{x^2}{x^{1/3} (x^3)^{1/6}} \\ &= \dfrac{x^2}{x^{1/3} x^{1/2}} \\ &= \dfrac{x^2}{x^{5/6}} \end{aligned} \]
Remember that we’ll need to get common denominators when adding the fractions at the end: \[ \begin{aligned} \frac{1}{3} + \frac{1}{2} &= \frac{1\times2}{3\times2} + \frac{1\times3}{2\times3} \\ &= \frac{2}{6} + \frac{3}{6} \\ &= \frac{5}{6} \end{aligned} \]
Finally, applying rule 4 and simplifying,
\[ \begin{aligned} \dfrac{x^2}{x^{5/6}} &=x^2 \times x^{-5/6} \\ &= x^{2-5/6} \\ &= x^{12/6-5/6} \\ &= x^{7/6} \end{aligned} \]
Lots of work with fractions here!
Now try these questions. Don’t worry if it takes a while to just solve one!
9.2.1 But why? Square roots
As promised here is an explanation of why \(x^{\frac{1}{n}} = \sqrt[n]x\).
When we take a square root we look for the a number that when it is multiplied by it’s self we get the answer i.e. \(? \times ? = x\). Since one \(x\) is the same as \(x^1\) we can rewrite out statement again:
\[ \begin{aligned} ? \times ? &= x^1 \\ x^? \times x^? &= x^1 \\ x^{?+?} &= x^1 \end{aligned} \] This means \(? + ? = 1\) so \(?=\frac{1}{2}\) so \(x^{\frac{1}{2}} = \sqrt{x}\).