# Indices and Surds

## Definition

An index (plural: indices) is the power, or exponent, of a number. For example, $a^3$ has an index of 3.

A surd is an irrational number that can be expressed with roots, such as $\sqrt{2}$ or $\sqrt[5]{19}$.

## Technique

The manipulation of indices and surds can be a powerful tool for evaluating and simplifying expressions.

\begin{aligned} x^m \times x^n &= x^{m+n} \\ \frac{x^m}{x^n} &= x^{m-n} \\ x^{-n} &= \frac{1}{x^n} \\ (x^m)^n &= x^{mn} \\ (xy)^{n} &= x^n y^n \\ \left(\frac{x}{y}\right)^m &= \frac{x^m}{y^m} \end{aligned}

Surds are just numbers with fractional indices, e.g. $\sqrt[3]{7^2} = 7^{2/3}$. Any operation with indices can be applied to surds, and indices and surds are related through this rule:

$x^{m/n} = \sqrt[n]{x^m}$

This allows us to group numbers together into forms that can be more convenient. Here are a couple examples:

### Simplify: $2^5 \times 4^3$

\begin{aligned} 2^5 \times 4^3 &= 2^5 \times (2^2)^3 \\ &= 2^5 \times 2^6 \\ &= 2^{11} \end{aligned} _\square

### Simplify: $\displaystyle \frac{(a^2)^4 b^7}{ab^{-2}}$

\begin{aligned} \frac{(a^2)^4 b^7}{ab^{-2}} &= \frac{a^8 b^7}{ab^{-2}} \\ &= \frac{a^7 b^7}{b^{-2}} \\ &= a^7 b^9 \end{aligned} _\square

Sometimes surds will appear in the dominator of an expression. You can rationalize the denominator by applying the following technique to a fraction of the form $\frac{a}{b+\sqrt{c}}$:

$\frac{a}{b+\sqrt{c}} \left( \frac{b-\sqrt{c}}{b-\sqrt{c}} \right) = \frac{ab-a\sqrt{c}}{b^2 -c}$ For example:

### Simplify: $\displaystyle \frac{1}{2+\sqrt{5}}$

\begin{aligned} \frac{1}{2+\sqrt{5}} &= \frac{1}{2+\sqrt{5}} \left( \frac{2-\sqrt{5}}{2-\sqrt{5}} \right) \\ &= \frac{2-\sqrt{5}}{2^2 -5} \\ &= -2 + \sqrt{5} \end{aligned} _\square

## Application and Extensions

### If you write the prime factorization of$\sqrt[5]{32^8}$, what is the sum of indices of the factors?

If you recognize that $32 = 2^5$, the answer falls quickly into place:

$\sqrt[5]{32^8} = (2^5)^\frac{8}{5} = 2^8$

So, the sum of the indices is simply 8. $_\square$

### If $3^{x-y} = 81$ and $3^{x+y} = 729$, what is $x$?

Multiply the two expressions together to get the $y$'s to cancel out: \begin{aligned} 3^{x-y} \times 3^{x+y} &= 81 \times 729 \\ 3^{x-y+x+y} &= 3^4 \times 3^6 \\ 3^{2x} &= 3^{10} \\ x &= 5 \end{aligned} _\square

Note by Arron Kau
6 years, 10 months ago

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How to solve this: (6^n+3-32.6^n+1)/(6^n+2-2.6^n+1)

- 6 years, 3 months ago

12.66

- 2 years, 4 months ago

Wrong

- 2 years, 1 month ago

Taking common out 6^(n+1) from num. & den. 6^2-32/6-2 = 1

- 1 year, 5 months ago

1

- 6 months, 2 weeks ago

0

- 5 months, 1 week ago