INDICES

An index (plural indices) shows how many times a number is multiplied by itself. It is also called a power or an exponent.

In the number a^n,\ a is called the base while n is the index. a^n is read as “a to the power of n.”

There are 2 basic rules you need to know with indices. This is simply how indices are defined for them to make sense matheematically:

  1. A number raised to a fraction whose numerator is one is equal to that root of the number. i.e  \large a^\frac{1}{n}=\sqrt[n]{a} .      e.g. \large a^\frac{1}{2}=\sqrt{a}\ ;\ a^\frac{1}{3}=\sqrt[3]{a}
  2. In a number raised to a fraction whose numerator is not one, the denominator is the root of the number while the numerator is the power of the number. i.e.  \large a^\frac{m}{n}=(\sqrt[n]{a})^m or  \large a^\frac{m}{n}=\sqrt[n]{a^m} .

Laws of Indices

There are a few laws that are used to manipulate exponents. Note that you do not need to know the names of these laws, only how to apply them.

Product Law: a^m\times a^n=a^{m+n}

When multiplying indices with the same base, add the powers.

To prove this, we’ll start with the left side of the equation and use the definition of exponentiation: a^m \times a^n = \underbrace{a \times a \times a \times \ldots \times a}_{m \text{ times}} \times \underbrace{a \times a \times a \times \ldots \times a}_{n \text{ times}} Now, consider combining these factors. = \underbrace{a \times a \times a \times \ldots \times a}_{m \text{ times}} \times \underbrace{a \times a \times a \times \ldots \times a}_{n \text{ times}} = \underbrace{a \times a \times a \times \ldots \times a}_{(m + n) \text{ times}} Now, we see that the expression on the right side is a^{m+n}, so we have: a^m \times a^n = a^{m+n}

Quotient Law: a^m\div a^n=a^{m-n}

When dividing indices with the same base, subtract the powers.

a^m \div a^n = \frac{a^m}{a^n} = \frac{\underbrace{a \times a \times \ldots \times a}_{m \text{ times}}}{\underbrace{a \times a \times \ldots \times a}_{n \text{ times}}} Now, consider canceling out common factors in the numerator and denominator. = \frac{\cancel{a} \times \cancel{a} \times \ldots \times a \times a \times \ldots}{\cancel{a} \times \cancel{a} \times \ldots \times \cancel{a}} = \underbrace{a \times a \times \ldots \times a}_{(m – n) \text{ times}} Now, we see that the expression on the right side is a^{m-n}, so we have: a^m \div a^n = a^{m-n}

Power Law: (a^m)^n=a^{m\times n}

When there is a power outside the bracket, multiply the powers.

(a^m)^n = a^{m \times n}To prove this, let’s start with the left side of the equation: =(\underbrace{a \times a \times \ldots \times a}_{m \text{ times}})^n Now, consider raising each factor to the power of n: =\underbrace{a^n \times a^n \times \ldots \times a^n}_{m \text{ times}} By the Product Law, we can combine these factors and obtain: =a^{n + n + \ldots + n}=a^{m \times n} Therefore, (a^m)^n = a^{m \times n}.

Negative exponent law: a^{-n}=\frac{1}{a^n}, and by extension, \displaystyle \left( \frac{a}{b} \right)^{-n} = \left(\frac{b}{a} \right)^n

A negative exponent indicates the reciprocal of the expression with the positive exponent. When a fraction is raised to a negative power, the reciprocal of the fraction will have a positive power.

By the quotient law, \frac{a^m}{ a^n}=a^{m-n} . In our case,  m=0. Thus we have: \frac{a^0}{ a^n}=a^{0-n} By the zero exponent law, a^0=1 . Thus we have: \frac{1}{a^n}=a^{-n}

Zero exponent law: a^0=1

Any non-zero value raised to the power of 0 is equal to 1.

By the quotient law, \frac{a^m}{ a^n}=a^{m-n} For the exponent to equal 0, the values of m and n have to be equal. We will set them equal to 1 for simplicity. \frac{a^1}{ a^1}=a^{1-1} \frac{a}{ a}=a^0 Thus we have: a^0=1

Example 1

Solve the equation 8^{x+1}-2^{3x-1}=120 .

First, we need to express the numbers to a common base in order to make the laws of indices apply. We can write 8 as 2^3 . Substituting this into the equation: \left( 2^3 \right)^{x+1}-2^{3x-1}=120 2^{3(x+1)}-2^{3x-1}=120 2^{3x+3}-2^{3x-1}=120 We now need to find a way to express 2^{3x+3} in terms of 2^{3x-1} . 2^{3x+3}=2^4\times2^{3x-1} Substituting this into the equation:
2^4 \left( 2^{3x-1} \right)-2^{3x-1}=120 Now, we can let 2^{3x-1} be y. 16y-y=120 15y=120 y=8 Thus, we have that 2^{3x-1}=8 . We can express 8 as 2^3 . 2^{3x-1}=2^3 Since the bases are equal, we can equate the exponents. 3x-1=3 3x=4 \boxed{x=1.333 }

 

Example 2

Evaluate: 27^{\frac{2}{3}}\times \left( \frac{81}{16} \right)^{-\frac{1}{4}}

We can approach this by first evaluating 27^{\frac{2}{3}}.  27^{\frac{2}{3}}= (\sqrt[3]{27})^2 (\sqrt[3]{27})^2=(3)^2 =9

Now, we evaluate \displaystyle\left( \frac{81}{16} \right)^{-\frac{1}{4}} . \left( \frac{81}{16} \right)^{-\frac{1}{4}}=\left( \frac{16}{81} \right)^{\frac{1}{4}} = \sqrt[4]{\frac{16}{81}} \sqrt[4]{\frac{16}{81}}=\frac{\sqrt[4]{16}}{\sqrt[4]{81}}= \frac{2}{3}

Our final expression is now  9\times\dfrac{2}{3}\ =\ 6 .

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