What is Exponent\ Power\ Index? || History of Exponent || Laws of Exponent

Exponent

• Before the definition of the exponent, we take an example to understand the importance of exponents
  
  

Find the prime factors of $256$

$256$ $=2$ $\times 2\times 2$ $\times 2\times 2$ $\times 2$ $\times 2$ $\times 2$

Repeated multiplication with the same number is lengthy, but if the number were long, the prime factors would also be long. Therefore, repeated multiplication needs to be written in the short form, and they write it as $2^8$.

Definition:

• The simplest approach to illustrate the repeated multiplication of an integer by itself is with exponents.
• How often a number (base) is multiplied by itself is shown by its exponent.
• The exponent is also called power or index.

$2^8$
$2$ is called the base of power and $8$ is called an exponent

$x^1$ $\times x^1$ $=x^2$

Each is an example of how polynomial $9x^4$ $+10x^3$ $+3x^2$ $+7x$ $+4$ was written in the past and present.

Some History

• Roman numerals were employed by Joost Burgi $(1552-1632)$ to denote an exponent above the coefficient.
  $9^{IV}$ $+10^{III}$ $+3^{II}$ $+7^{I}+4$

• Adriann Van Roomen $(1561-1615)$ used parenthesis
  $9(4)$ $+10(3)$ $+3(2)$ $+7(1)$ $+4$

• Pierre Herigone $(1580-1643)$ placed the coefficient before the variable and the exponent after.
  $9x4$ $+10x3$ $+3x2$ $+7x1$ $+4$

• Roman numerals were used by James Hume to represent an increased exponent.
  $9x^{iv}$ $+10x^{iii}$ $+3x^{ii}$ $+7x^{i}$ $+4$

• Rene Descartes $(1596-1650)$ introduced the symbolism that is commonly used today.
  $9x^{4}$ $+10x^{3}$ $+3x^{2}$ $+7x$ $+4$

$3^5$  (read as “$3$ to the power $5$”)

(or the $5^{th}$ power of $3$)

$3^2$ $=\underset{2\text{ }factors}{3\times 3 }$

$(-2)^2$ $=\underset{4\text{ }factors}{(-2)\times (-2)\times (-2)\times (-2) }$

$(8)^5$ $=\underset{5\text{ }factors}{8\times 8\times 8\times 8\times 8 }$

$\left( \frac{7}{9} \right)^6$ $=\frac{7}{9}$ $\times$ $\frac{7}{9}$ $\times$ $\frac{7}{9}$ $\times$ $\frac{7}{9}$ $\times$ $\frac{7}{9}$ $\times$ $\frac{7}{9}$

This idea may be extended to products of $n$ components, each of which is represented by $"a"$ in the equation.

$a^n$ $=\underset{n\text{ factor of }a}{a\times a\times a\times.......a}$  where  $“n”$ is a natural number $n\epsilon N$ $\wedge$ $a\epsilon$ $\mathbb{R}$

In the symbol $a^n$, the $nth$ power of $a$, $“a”$ is called the base and $“n”$ is the exponent or index of the base.

In  $3^{-4}=$ $\frac{1}{3^{4}}$

$=\left( \frac{1}{4} \right)^4$

$\frac{1}{3}$ is the base and $4$ is the exponent

OR
$3$ is the base and $-4$ is the exponent

It may be noted that:

$(-3)^4$ $=\left( -3 \right)$ $\times$ $\left( -3 \right)$ $\times$ $\left( -3 \right)$ $\times$ $\left( -3 \right)$ $=81$

And
$-(3)^4$ $=-\left( 3 \right)$ $\left( 3 \right)$ $\left( 3 \right)$ $\left( 3 \right)$ $=-81$

For the sake of convenience, we write

$-\left( 3 \right)^4$ as $-3^4$ and

$-\left( a \right)^n$ is generally written as $-a^n$

Example $(i)$: Find the value of $2\left( 3 \right)^5$

$2\left( 3 \right)^5$ $=2\left( 3\times 3\times 3\times 3\times 3\right)$

$=2\left( 243\right)$

$=486$

Keep in mind the outcomes listed below.

• If $a$ is a positive real number, then $a^n$ is positive, $\forall$, $n\epsilon$ $\mathbb{N}$

Example:
$(i)$  $(5)^3$ $=125$,  $(0.5)^3$ $=0.125$,  $(0.4)^2$ $=0.16$

$(ii)$  If $a$ is a negative real number and $n$ is even, then $a^n$ is positive.

Example:

$(-a)^n$ $=a^n$, $n$ is even

$(-5)^4$ $=5^4$ $=625$

$(-2)^6$ $=2^6$ $=64$

$(iii)$  If $a$ is a negative real number and $n$ is odd, then $a^n$ is negative.

Example: $(-a)^n$ $=-a^n$, $n$ is odd

$(-4)^3$ $=-4^3$ $=-64$

$(-3)^3$ $=-27$

$(-5)^5$ $=-3725$

Laws of Exponent

$1)$  Product Law/Multiplication Law

Example:

$(i)$  $5^3$ $\times 5^2$

Solution:

$=(5$ $\times 5$ $\times 5$ $\times 5)$

$=5^{3+2}$

$(ii)$  $a^3$ $\times a^4$

Solution:

$=(a\times a$ $\times a)$ $\times (a$ $\times a$ $\times a$ $\times a)$

$=a$ $\times a\times $ $a$ $\times a$ $\times a\times a$ $\times a$

$=a^7$

$=a^{4+3}$

From the above examples, we deduce that,
When exponents are added to a product with several bases of various powers, the base does not change.
Thus, $a^m$ $\times a^n$ $=a^{m+n}$

where $"a"$ is a real number and $m$, $n$ are natural numbers
Similarly,  $a^m$ $\times a^n$ $\times a^p$ $=a^{m+n+p}$, where $m$, $n$, and $p$ are natural numbers

Example: Simplify: $a^5$ $\times b^3$ $\times c^8$ $\times b^{10}$ $\times a^{13}$ $\times c^4$

Solution:
$a^5$ $\times b^3$ $\times c^8$ $\times b^{10}$ $\times a^{13}$ $\times c^4$

$(a^5$ $\times a^{13})$ $\times (b^3\times b^{10})$ $\times (c^8\times c^4)$

$=a^{5+13}$ $\times b^{3+10}$ $\times c^{8+4}$

$=a^{18}$ $b^{13}$ $c^{12}$

2.    Law of power of a product

$(a\times b)^5$ $=(a\times b)$ $\times (a$ $\times b)$ $\times (a$ $\times b)$ $\times (a$ $\times b)$ $\times (a$ $\times b)$

$=\left(a\times a \times a \times a\times a\right)$ $\times$ $(b$ $\times b$ $\times b$ $\times b$ $\times b)$

$=a^5$ $\times b^5$

From this, we deduce that the power of the product of two numbers is equal to the product of their powers.

Thus
$(a.b)^n$ $=a^n$ . $b^n$

Whereas $n$ is a natural number and $a$ and $b$ are real numbers.
Similarly, it can be shown that
$(abc)^n$ $=a^n$ $.b^n$ $.c^n$, where $a$, $b$ and $c$ are real numbers

$3)$  Law of the power of a power
Consider the following examples
$(3^2)^4$ $=3^2$ $\times 3^2$ $\times 3^2$ $\times 3^2$

Consider the following examples

$(3^2)^4$ $=3^2$ $\times 3^2$ $\times 3^2$ $\times 3^2$

$=3^{2+2+2+2}$

$=3^8$

$=3^{2\times 4}$

$(a^5)^3$

$=a^5$ $\times a^5$ $\times a^5$

$=a^{5+5+5}$

$=a^{15}$

$=a^{5\times 3}$

$(b^7)^4$

$=b^7$ $\times b^7$ $\times b^7$ $\times b^7$

$=b^{7+7+7+7}$ $=b^{28}$

$=b^{7\times 4}$

From these examples, we deduce that

Keep the base constant and double the exponents when increasing power to a power.

$(a^m)^n$ $=a^{mn}$
where, $m$, $n\epsilon \mathbb{Z}$, $a$ is a real number
so,
$i)$  $(6^3)^6$

$=6^{3\times 6}$

$=6^{18}$

$ii)$  $(5^2)^7$

$=5^{2\times 7}$

$=5^{14}$

$iii)$  $\left[ \left( -5 \right)^2 \right]^8$

$=\left[ 5^2 \right]^8$

$=5^{2\times 8}$

$=5^{16}$

It can be shown that
$\left[ (a^m)^n \right]^p$ $=a^{mnp}$

$4)$  Law of Quotient of powers

Let us simplify the following expressions

$i)$  $\frac{a^7}{a^3}$

$=\frac{a\times a\times a\times a\times a\times a\times a}{a\times a\times a}$

$=a\times a\times a\times a$

$=a^4$

$=a^{7-3}$

$ii)$  $\frac{(-8)^6}{(-8)^4}$

$=\frac{(-8)(-8)(-8)(-8)(-8)(-8)}{(-8)(-8)(-8)(-8)}$

$=(-8)(-8)$

$=(-8)^2$

$=(-8)^{6-4}$

$iii)$  $\frac{a^n}{a^m}$

$=\frac{\underset{\text{n factors}}{a\times a\times a....a}}{\underset{\text{m factors}}{a\times a\times a....a}}$

We deduce that
When a quotient of powers has the same base, we can maintain the base by subtracting the exponents of the denominator and numerator.

Thus,
$\frac{a^m}{a^n}$

$=a^{m-n}$
where $a$ is a non-zero real number
i.e $a\neq 0$ and $m$, $n$ are integers

Note: $(i)$ By the above law if $m=n$, then

$\frac{a^n}{a^n}$ $=a^{n-n}$ $=a^0$,    $a\neq 0$

But we know that

$\frac{a^n}{a^n}$

$=1$
Hence
$a^0=1$

Where $a$ is a non-zero real number

Example:
$(1001)^0$ $=1$,    $(\sqrt{2})^0$ $=1$,    $(-5)^0$ $=1$
$(0.25)^0$ $=1$,     $(0.0002)^0$ $=1$

Any non-zero number raised to the power of zero is $1$.

$(ii)$  $a^n\times a^{-n}$ $=1$,        $a\neq 0$

Solution:

$a^n\times a^{-n}$

$=a^{n-n}$

$=a^0$ $=1$

$a^{-n}$ $=\frac{1}{a^n}$ and  $a^{n}$ $=\frac{1}{a^{-n}}$

Thus,
$a^{-1}$ $=\frac{1}{a}$  and $a^{-2}$ $=\frac{1}{a^2}$

Simplify
$(i)$  $\frac{(m+n)^7(p+q)^5}{(m+n)^6(p+q)^2}$

Solution:
$(m+n)^{7-6}(p+q)^{5-2}$

$(m+n)(p+q)^3$

$5.$  Law of power of a Quotient
Let us simplify the following expressions

$i)$  $\left( \frac{7}{4} \right)^4$

$=\left( \frac{7}{4} \right)$ $\left( \frac{7}{4} \right)$ $\left( \frac{7}{4} \right)$ $\left( \frac{7}{4} \right)$

$=\frac{7\times 7\times 7\times 7}{4\times 4\times 4\times 4}$

$=\frac {7^4}{4^4}$

$ii)$  $\left( \frac{6}{7} \right)^5$

$=\left( \frac{6}{7} \right)$ $\left( \frac{6}{7} \right)$ $\left( \frac{6}{7} \right)$ $\left( \frac{6}{7} \right)$ $\left( \frac{6}{7} \right)$

$=\frac{6\times 6\times 6\times 6\times 6}{7\times 7\times 7\times 7\times 7}$

$=\frac{6^5}{7^5}$

Similarly, for any rational number $n$.

$\left( \frac{a}{b} \right)^n$

$=\left( \frac{a}{b} \right)$ $\left( \frac{a}{b} \right)$ $\left( \frac{a}{b} \right)$ $\times ....$ $\times $ $\left( \frac{a}{b} \right)$ $\text{ ( n times )}$

$=\frac{a\times a\times a\times....\times a\text{ (n times)}}{b\times b\times b\times....\times b\text{ (n times)}}$ $=\frac{a^n}{b^n}$,  $n\epsilon \mathbb{N}$

For any two real numbers $"a"$ and $"b"$ $($ where $b\neq 0)$, $n$ being a natural number.
$\left( \frac{a}{b} \right)^n$ $=\frac{a^n}{b^n}$

This is called the law of the power of a quotient

Note: $\left( \frac{a}{b} \right)^{-n}$ $=\frac{a^{-n}}{b^{-n}}$

$=\frac{b^n}{a^n}$     $($ since $a^{-n}$ $=\frac{1}{a^n})$

$\left( \frac{a}{b} \right)^{-1}$ $=\left( \frac{b}{a} \right)^n$

Example:

$\left( \frac{-30x^{10}y^8}{-5x^3y^2} \right)^2$

$=\left(\frac{-30}{-5}\left(x^{10-3}y^{8-2} \right)\right)^2$

$=\left( 6\left( x^7y^6 \right) \right)^2$

$=36x^{14}y^{12}$