Definition of Exponent Numbers
The following positive integer numbers ($ n in Z^+ $), $ a $ is called the base number or principal number and $ n $ is called the exponent defined as follows.
$ a^n = underbrace {a times a times a times … times a}_{mbox{n factor}} $; $ a in R $
Whereas the negative exponent is defined as follows.
$ begin{align} a^{- m} & = (frac{1}{a})^m \ & = frac{1}{a^m} end{align} $; $ a in R $
Example:
$ 2^3 = 2 times 2 times 2 = 8 $
$ (- 4)^3 = (- 4) times (-4) times (-4) = – 64 $
$ 2^{-3} = frac{1}{2^3} = frac{1}{8} $
Zero Exponent
Not all real numbers raised by 0 are 1. However, there are exceptions that this does not apply to 0. So, for every $ a in R $; $ a neq 0 $ applies:
$ a^0 = 1 $
Evidence:
$ begin{align} frac{a^n}{a^n} & = 1 \ Leftrightarrow a^{n-n} & = 1 \ Leftrightarrow a^{0} & = 1 end{align} $
Characteristics of Positive Exponent
If $ a in R $, m and n are natural numbers then
$ a^m times a^n = a^{m + n} $
$ a^m:a^n = a^{m-n} $
$ (a^m)^n = a^{mn} $
$ (a times b)^n = a^n b^n $
$ (frac{a}{b})^n = frac{a^n}{b^n}; b neq 0 $
Characteristics of Fraction Exponent
$ (a^{frac{m}{n}})(a^{frac{p}{n}}) = a^{frac{m + p}{n}} $
$(a^{frac{m}{n}})(a^{frac{p}{q}}) = a^{frac{m}{n} + frac{p}{q}} $
Sample Questions
Examples of school questions:
1. $ 2^2 times 2^5 = 2^{2 + 5} = 2^{7} $
2. $ (2^3)^2 = 2^6 $
3. $ (frac{2}{3})^3 = frac{2^3}{3^3} = frac{8}{27} $
The following numbers whose biggest value is …
a. 777
b. $ 7^{77} $
c. $ (77)^7 $
d. $ (7^7)^7 $
e. $ (7 times 7)^7 $
Answer B
We use the properties of a number to answer the question.
a. $ 777 = 7(111) <7(11^2) $
b. $ 7^{77} = 7^{57} 7^{18} 7^2 = 7^{57} 7^{18} 49 $
c. $ 77^7 = (7 times 11)^7 = 7^7 times 11^7 = 7^7 11^611 $
d. $ (7^7)^7 = 7^{49} $
e. $ (7 times 7)^7 = (7^2)^7 = 7^{14} $
Note that $ 7^{77} > 7^{49} > 7^{14}$, therefore $ b > d > e $. Because $ 7^3 > 11^2 $ then $ 7^{57} 7^{18} 49 > 7^7 11^6 11 > 7 times 11^2 > 7(111) $, so $ b > c> a $.
Examples of Olympic questions: Determine the unit number of $ 7^{1234}$
Solution:
Consider the following pattern!
$ begin{align} 7^1 & = 7 \ 7^{2} & = … 9 \ 7^{3} & = … 3 \ 7^{4} & = … 1 \ 7^{ 5} & = … 7 \ 7^{6} & = … 9 \ 7^{7} & = … 3 \ 7^{8} & = … 1 end{align} $
Based on this pattern and using the properties of the Exponent numbers obtained:
$ begin{align} 7^{1234} & = 7^{(4 times 308)} times 7^2 \ & = (7^4)^{308} times 7^2 end{align}$.
Because units of $ 7^4 $ are 1 and $ 7^2 $ is 9, the unit number of $ 7^{1234} $ is $ 1 times 9 = 9 $.