國中數學/國中數學七年級/2-5 指數律

在舊課綱中，該單元合併在1-4 指數記法與科學記號。但在108新課綱後，該單元改為併入2-3 分數的加減及2-4 分數的乘除。但因其重要性，因此將本單元獨立成一個單元。

指數律

$a,b,m,n,x$是不為$0$的實數($a,b,m,n,x\in ℝ\ne 0$)，且a,b不可同時為負數

a^m × aⁿ = a^m+n

$\begin{array}{cc}& a^m\times a^n\\ =& (\underset{m\text{個}a}{\underbrace{a\times a\times a\times \cdots \times a}})\times (\underset{n\text{個}a}{\underbrace{a\times a\times \cdots \times a}})\\ =& \underset{m\text{個}a}{\underbrace{a\times a\times a\times \cdots \times a}}\times \underset{n\text{個}a}{\underbrace{a\times a\times \cdots \times a}}\\ =& \underset{m+n\text{個}a}{\underbrace{a\times a\times a\times \cdots \times a}}\\ =& a^{m+n}\end{array}$

a^m ÷ aⁿ = a^m-n

$\begin{array}{cc}& \text{(1)}\\ & \text{if}m>n:\\ & a^m÷a^n\\ =& \frac{\stackrel{m\text{個}a}{\overbrace{a\times a\times a\times \cdots \times a}}}{\underset{n\text{個}a}{\underbrace{a\times a\times \cdots \times a}}}\\ =& \frac{\stackrel{m\text{個}a}{\overbrace{a\times a\times a\times \cdots ̸\times a\times a\times \cdots \times a}}}{\underset{n\text{個}a}{\underbrace{a\times a\times \cdots ̸\times a}}}\\ =& \underset{m-n\text{個}a}{\underbrace{a\times a\times a\cdots a}}\\ =& a^{m-n}\\ & \\ & \text{(2)}\\ & \text{if}m<n:\\ & a^m÷a^n\\ =& \frac{\stackrel{m\text{個}a}{\overbrace{a\times a\times \cdots \times a}}}{\underset{n\text{個}a}{\underbrace{a\times a\times a\times \cdots \times a}}}\\ =& \frac{\stackrel{m\text{個}a}{\overbrace{a\times a\times \cdots ̸\times a}}}{\underset{n\text{個}a}{\underbrace{a\times a\times a\times \cdots ̸\times a\times a\times \cdots \times a}}}\\ =& \frac{1}{\underset{n-m\text{個}a}{\underbrace{a\times a\times a\cdots a}}}\\ =& \frac{1}{a^{n-m}}\end{array}$

a⁰ = 1

$\begin{array}{cc}& \text{if}m=n:\\ & a^m÷a^n\\ =& a^{m-n}=a^0\\ =& \frac{\stackrel{m\text{個}a}{\overbrace{a\times a\times a\times \cdots \times a}}}{\underset{n\text{個}a}{\underbrace{a\times a\times a\times \cdots \times a}}}\\ =& \frac{\stackrel{m\text{個}a}{\overbrace{a\times a\times \cdots ̸\times a}}}{\underset{n\text{個}a}{\underbrace{a\times a\times \cdots ̸\times a}}}\\ =& 1\\ \Rightarrow & a^0=1\end{array}$

(a^m)ⁿ = a^m×n

$\begin{array}{cc}& (a^m{)}^n\\ =& (\underset{n\text{個}{a}^m}{\underbrace{a^m\times a^m\times a^m\times \cdots \times a^m}})\\ =& \underset{m\times n\text{個}a}{\underbrace{\stackrel{m}{\overbrace{(a\times \cdots \times a)}}\times \stackrel{m}{\overbrace{(a\times \cdots \times a)}}\times \cdots \times \stackrel{m}{\overbrace{(a\times \cdots \times a)}}}}\\ =& \underset{m\times n\text{個}a}{\underbrace{a\times a\times a\times \cdots \times a}}\\ =& a^{m\times n}\end{array}$

a^m × b^m = (a × b)^m

$\begin{array}{cc}& a^m\times b^m\\ =& (\underset{m\text{個}a}{\underbrace{a\times a\times \cdots \times a}})\times (\underset{m\text{個}b}{\underbrace{b\times b\times \cdots \times b}})\\ =& \underset{m\text{個}a}{\underbrace{a\times a\times \cdots \times a}}\times \underset{m\text{個}b}{\underbrace{b\times b\times \cdots \times b}}\\ =& \underset{m\text{個}a\text{和}b\text{相 乘}}{\underbrace{a\times b\times a\times b\times a\times b\times a\times b\times \cdots \times a\times b}}\\ =& \underset{m\text{個}a\times b}{\underbrace{(a\times b)\times (a\times b)\times (a\times b)\times (a\times b)\times \cdots \times (a\times b)}}\\ =& (a\times b{)}^m\end{array}$

a^m ÷ b^m = (a ÷ b)^m

$\begin{array}{cc}& a^m÷b^m\\ =& (\underset{m\text{個}a}{\underbrace{a\times a\times \cdots \times a}})÷(\underset{m\text{個}b}{\underbrace{b\times b\times \cdots \times b}})\\ =& \underset{m\text{個}a}{\underbrace{a\times a\times \cdots \times a}}÷\underset{m\text{個}b}{\underbrace{b÷b÷\cdots ÷b}}\\ =& \underset{m\text{個}a\text{和}b\text{相 除}}{\underbrace{a÷b\times a÷b\times a÷b\times a÷b\times \cdots \times a÷b}}\\ =& \underset{m\text{個}a÷b}{\underbrace{(a÷b)\times (a÷b)\times (a÷b)\times (a÷b)\times \cdots \times (a÷b)}}\\ =& (a÷b{)}^m\end{array}$

(^b⁄_a)^m = ^bm⁄_a^m

$\begin{array}{cc}& (\frac{a}{b}{)}^m\\ =& (a÷b{)}^m\\ =& a^m÷b^m\\ =& \frac{a^m}{b^m}\end{array}$

a^-x = ¹⁄_a^x

$\begin{array}{cc}& a^{-x}\\ =& (a^{-1}{)}^x\\ =& (a^{m-n}{)}^x(m-n=-1)\\ =& (a^m-a^n{)}^x\\ =& (\frac{\stackrel{m\text{個}a}{\overbrace{a\times a\times \cdots \times a}}}{\underset{n\text{個}a}{\underbrace{a\times a\times a\times \cdots \times a}}}{)}^x\\ =& \frac{(\stackrel{m\text{個}a}{\overbrace{a\times a\times \cdots \times a}}{)}^x}{(\underset{n\text{個}a}{\underbrace{a\times a\times a\times \cdots \times a}}{)}^x}\\ =& \frac{\stackrel{m\text{個}{a}^x}{\overbrace{a^x\times a^x\times \cdots \times a^x}}}{\underset{n\text{個}{a}^x}{\underbrace{a^x\times a^x\times a^x\times \cdots \times a^x}}}\\ =& \frac{\stackrel{m\text{個}a}{\overbrace{a^x\times a^x\times \cdots ̸\times a^x}}}{\underset{n\text{個}{a}^x}{\underbrace{a^x\times a^x\times a^x\times \cdots ̸\times a^x}}}\\ =& \frac{1}{\underset{n-m=1\text{個}{a}^x}{\underbrace{a^x}}}\\ =& \frac{1}{a^x}\end{array}$

習題

1.$(10\times 5^2)÷(2\times 5^{-3})=?$
(A) 1 (B) 5⁴ (C) 5⁵ (D) 5⁶
2.$(-4^4{)}^2÷2^{10}÷8^2$
(A) 1 (B) 2 (C) 4 (D) 8

習題解答及解析

1.(D) 2.(A)
1.
$\begin{array}{cc}& (10\times 5^2)÷(2\times 5^{-3})\\ =& 10\times 5^2÷2÷5^{-3}\\ =& (10÷2)\times (5^2÷5^{-3})\\ =& 5\times 5^{2-(-3)}\\ =& 5^1\times 5^5\\ =& 5^6\end{array}$
2.
$\begin{array}{cc}& (-4^4{)}^2÷2^{10}÷8^2\\ =& 2^{16}÷2^{10}÷2^6\\ =& 2^{(16-10-6)}\\ =& 2^0\\ =& 1\end{array}$