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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.

Let's start with some basic rules for operations with indices:

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

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: \( 2^5 \times 4^3 \)

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

 

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

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

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: \( \displaystyle \frac{1}{2+\sqrt{5}} \)

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

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: \[ \begin{align} 3^{x-y} \times 3^{x+y} &= 81 \times 729 \\
3^{x-y+x+y} &= 3^4 \times 3^6 \\
3^{2x} &= 3^{10} \\ x &= 5 \end{align} _\square \]

Note by Arron Kau
3 years, 3 months ago

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How to solve this: (6^n+3-32.6^n+1)/(6^n+2-2.6^n+1) Shilpa Negandhi · 2 years, 9 months ago

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