# 高中數學/函數與三角/對數的概念與運算

## 基礎知識

### 定義

• 符號${\displaystyle \log _{a}N}$表示有多少個a連續相乘會等於N，或者說a的多少次冪會等於N。
• 沒有以零或以負數為真數N的對數，或者說它們不能作為合法的真數，對它們無法施加取對數值的運算。因為高中學習的是實變量的函數，由${\displaystyle a>0,b\in \mathbb {R} }$可知一定有${\displaystyle a^{b}>0}$，所以在實數範圍並不存在使得真數取零或取負數值的情形。[1]
• ${\displaystyle \log _{a}1=0,\log _{a}a=1\quad (a>0,a\neq 1)}$[1]

• 通常將以10為底數的對數叫做常用對數。為了書寫簡便，N的常用對數${\displaystyle \log _{10}N}$簡記作${\displaystyle \lg N}$
• 通常將以特殊無理數${\displaystyle e=2.71828...}$為底數的對數叫做常用對數。為了書寫簡便，N的常用對數${\displaystyle \log _{e}N}$簡記作${\displaystyle \ln N}$

(1) ${\displaystyle \log _{25}5}$
(2) ${\displaystyle \lg {\frac {1}{10}}}$
(3) ${\displaystyle \ln e^{5}}$；(4) ${\displaystyle \lg 1}$

### 運算規律

• 和差關係：${\displaystyle \log _{a}MN=\log _{a}M+\log _{a}N}$${\displaystyle \log _{a}{\frac {M}{N}}=\log _{a}M-\log _{a}N}$

• 基變換（換底公式）：${\displaystyle \log _{a}b={\frac {\log _{c}b}{\log _{c}a}}}$

• 指係（次方公式）：${\displaystyle \log _{a^{n}}b^{m}={\frac {m}{n}}\log _{a}b}$

${\displaystyle \log _{a}{\frac {M}{N}}=\log _{a}M-\log _{a}N}$

(1) ${\displaystyle \log _{a}2+\log _{a}{\frac {1}{2}}}$
(2) ${\displaystyle \log _{3}18-\log _{3}2}$
(3) ${\displaystyle \lg {\frac {1}{4}}-\lg 25}$
(4) ${\displaystyle 2\log _{5}10+\log _{5}0.25}$
(5) ${\displaystyle 2\log _{5}25+3\log _{2}64}$
(6) ${\displaystyle \log _{2}(\log _{2}16)}$

(1) ${\displaystyle \log _{a}{\frac {\sqrt[{3}]{x}}{y^{2}z}}}$
(2) ${\displaystyle \log _{a}(x{\sqrt[{4}]{\frac {z^{3}}{y^{2}}}})}$
(3) ${\displaystyle \log _{a}(xy^{\frac {1}{2}}z^{-{\frac {2}{3}}})}$
(4) ${\displaystyle \log _{a}{\frac {xy}{x^{2}-y^{2}}}}$
(5) ${\displaystyle \log _{a}({\frac {x+y}{x-y}}\times y)}$
(6) ${\displaystyle \log _{a}({\frac {y}{x(x-y)}})^{3}}$

(1) ${\displaystyle e^{\ln 3}+\log _{\sqrt {5}}25}$
(2) ${\displaystyle 2^{\ln e+\lg 1}+3^{\log _{3}2}+3^{\log _{3}4-\lg 10}+2^{\ln 1}}$
(3) ${\displaystyle 2\log _{3}2-\log _{3}{\frac {32}{9}}+\log _{3}8-25^{\log _{5}3}}$
(4) ${\displaystyle (\log _{2}5+\log _{4}0.2)\times (\log _{5}2-\log _{25}0.5)}$
(5) ${\displaystyle \lg 125+\lg 2\lg 500+(\lg 2)^{2}}$
(6) ${\displaystyle {\frac {(1-\log _{6}3)^{2}+\log _{6}2\times \log _{6}18}{\log _{6}4}}}$

(1)
${\displaystyle {\begin{array}{l}e^{\ln 3}+\log _{\sqrt {5}}25=3+\log _{\sqrt {5}}({\sqrt {5}})^{4}\\=3+4\log _{\sqrt {5}}{\sqrt {5}}\\=3+4=7\end{array}}}$
(2)
${\displaystyle {\begin{array}{l}2^{\ln e+\lg 1}+3^{\log _{3}2}+3^{\log _{3}4-\lg 10}+2^{ln1}\\=2^{1+0}+2+{\frac {3^{\log _{3}4}}{3^{\lg 10}}}+2^{0}\\=2+2+{\frac {4}{3^{1}}}+1\\=5+{\frac {4}{3}}={\frac {19}{3}}\end{array}}}$
(3)
${\displaystyle {\begin{array}{l}2\log _{3}2-\log _{3}{\frac {32}{9}}+\log _{3}2^{3}-25^{\log _{5}3}\\=2\log _{3}2-\log _{3}{\frac {2^{5}}{3^{2}}}+3\log _{3}2-5^{2\log _{5}3}\\=2\log _{3}2-(\log _{3}2^{5}-\log _{3}3^{2})+3\log _{3}2-5^{\log _{5}3^{2}}\\=2\log _{3}2-(5\log _{3}2-2\log _{3}3)+3\log _{3}2-3^{2}\\=2\log _{3}2-(5\log _{3}2-2)+3\log _{3}2-9\\=2\log _{3}2-5\log _{3}2+2+3\log _{3}2-9\\=(2\log _{3}2-5\log _{3}2+3\log _{3}2)+(2-9)\\=0+(-7)=-7\end{array}}}$
(4)
${\displaystyle {\begin{array}{l}(\log _{2}5+\log _{2^{2}}0.2)\times (\log _{5}2-\log _{25}0.5)\\=(\log _{2}5+\log _{2^{2}}5^{-1})\times (\log _{5}2-\log _{5^{2}}2^{-1})\\=(\log _{2}5+{\frac {-1}{2}}\log _{2}5)\times (\log _{5}2-{\frac {-1}{2}}\log _{5}2)\\=(\log _{2}5-{\frac {1}{2}}\log _{2}5)\times (\log _{5}2+{\frac {1}{2}}\log _{5}2)\\={\frac {1}{2}}\log _{2}5\times {\frac {3}{2}}\log _{5}2\\=({\frac {1}{2}}\cdot {\frac {3}{2}})\cdot (\log _{2}5\log _{5}2)\\={\frac {3}{4}}\cdot 1={\frac {3}{4}}\end{array}}}$
(5)
${\displaystyle {\begin{array}{l}\lg 125+\lg 2\lg 500+(\lg 2)^{2}\\=\lg(5^{3})+\lg 2\lg(5\times 10^{2})+(\lg 2)^{2}\\=3\lg 5+\lg 2(\lg 5+\lg 10^{2})+(\lg 2)^{2}\\=3\lg 5+\lg 2(\lg 5+2\lg 10)+(\lg 2)^{2}\\=3\lg 5+\lg 2(\lg 5+2)+(\lg 2)^{2}\\=3\lg 5+(\lg 2\lg 5+2\lg 2)+(\lg 2)^{2}\\=3\lg 5+2\lg 2+(\lg 2\lg 5+(\lg 2)^{2})\\=3\lg 5+2\lg 2+(\lg 5+\lg 2)\lg 2\\=3\lg 5+2\lg 2+\lg(5\times 2)\lg 2\\=3\lg 5+2\lg 2+\lg 10\lg 2\\=3\lg 5+2\lg 2+\lg 2\\=3(\lg 5+\lg 2)\\=3\lg(5\times 2)\\=3\lg 10=3\end{array}}}$
(6)
${\displaystyle {\begin{array}{l}{\frac {(1-\log _{6}3)^{2}+\log _{6}2\times \log _{6}18}{\log _{6}4}}\\={\frac {(1-\log _{6}3)^{2}+\log _{6}2\times \log _{6}(2\times 3^{2})}{\log _{6}2^{2}}}\\={\frac {(1-\log _{6}3)^{2}+\log _{6}2\times (\log _{6}2+\log _{6}3^{2})}{2\log _{6}2}}\\={\frac {(1-\log _{6}3)^{2}+\log _{6}2\times (\log _{6}2+2\log _{6}3)}{2\log _{6}2}}\\={\frac {(1-\log _{6}3)^{2}+(\log _{6}2)^{2}+2\log _{6}2\log _{6}3}{2\log _{6}2}}\\={\frac {1-2\log _{6}3+(\log _{6}3)^{2}+(\log _{6}2)^{2}+2\log _{6}2\log _{6}3}{2\log _{6}2}}\\={\frac {1-2\log _{6}3+((\log _{6}3)^{2}+(\log _{6}2)^{2}+2\log _{6}2\log _{6}3)}{2\log _{6}2}}\\={\frac {1-2\log _{6}3+(\log _{6}3+\log _{6}2)^{2}}{2\log _{6}2}}\\={\frac {1-2\log _{6}3+(\log _{6}(3\times 2))^{2}}{2\log _{6}2}}\\={\frac {1-2\log _{6}3+(\log _{6}6)^{2}}{2\log _{6}2}}\\={\frac {1-2\log _{6}3+1^{2}}{2\log _{6}2}}\\={\frac {2(1-\log _{6}3)}{2\log _{6}2}}\\={\frac {1-\log _{6}3}{\log _{6}2}}\\={\frac {1}{\log _{6}2}}-{\frac {\log _{6}3}{\log _{6}2}}\\=\log _{2}6-\log _{2}3\\=\log _{2}{\frac {6}{3}}\\=\log _{2}2=1\\\end{array}}}$

${\displaystyle \log _{a}M^{\log _{a}N}=\log _{a}N^{\log _{a}M}\quad \Leftrightarrow \quad \log _{a}N\log _{a}M=\log _{a}M\log _{a}N}$

${\displaystyle a\ln a=b\ln b}$帶入上式可得：
${\displaystyle {\frac {a}{b}}-{\frac {3b}{a}}=2\quad \Rightarrow \quad a^{2}-3b^{2}=2ab\quad \Rightarrow \quad a^{2}-2ab+b^{2}-4b^{2}=0\quad \Rightarrow \quad (a-b)^{2}=4b^{2}\quad \Rightarrow \quad a-b=\pm 2b}$

${\displaystyle 3b\ln(3b)=b\ln b\quad \Rightarrow \quad (3b)^{3}=b\quad \Rightarrow \quad b={\frac {\sqrt {3}}{9}}}$

(1) 求${\displaystyle a^{m+2n}}$的值。
(2) 若${\displaystyle 0，求${\displaystyle x^{2}+x^{-2}}$的值。

(1)由${\displaystyle log_{a}3=m}$可知${\displaystyle a^{m}=3}$，由${\displaystyle \log _{a}2=n}$可知${\displaystyle a^{n}=2}$
${\displaystyle a^{m+2n}=a^{m}a^{2n}=a^{m}(a^{n})^{2}=3\times (2)^{2}=12}$
(2)${\displaystyle m+n=\log _{3}2+1\quad \Rightarrow \quad \log _{a}3+\log _{a}2=\log _{3}2+\log _{3}3\quad \Rightarrow \quad \log _{a}(3\times 2)=\log _{3}(3\times 2)\quad \Rightarrow \quad a=3}$

## 補充習題

• 已知a、b、c是三角形ABC的3條邊，且關於x的二次方程${\displaystyle x^{2}-2x+\lg(c^{2}-b^{2})-2\lg a+1=0}$有2個相等的實數根，則此三角形的形狀是(    )。
A.銳角三角形；B.直角三角形；C.鈍角三角形；D.等邊三角形
• 計算${\displaystyle (\lg 2)^{2}+\lg 5\times \lg 20-1}$

{\displaystyle {\begin{aligned}&(\lg 2)^{2}+\lg 5\times \lg 20-1\\&=(\lg 2)^{2}+\lg 5\times \lg(4\times 5)-1\\&=(\lg 2)^{2}+\lg 5\times (\lg 4+\lg 5)-1\\&=(\lg 2)^{2}+\lg 4\times \lg 5+(\lg 5)^{2}-1\\&=(\lg 2)^{2}+\lg 2^{2}\times \lg 5+(\lg 5)^{2}-1\\&=(\lg 2)^{2}+2\lg 2\times \lg 5+(\lg 5)^{2}-1\\&=(\lg 2+\lg 5)^{2}-1=(\lg(2\times 5))^{2}-1\\&=(\lg 10)^{2}-1=1^{2}-1=0\end{aligned}}}

• 已知芮氏地震等級R與地震波釋放的能量E的關係為${\displaystyle R={\frac {2}{3}}(\lg E-11.4)}$，求9級地震釋放的能量是8級地震的釋放能量的多少倍？

## 參考資料

1. 人民教育出版社中學數學室. 第2章「函數」第2.7節「對數」. 數學. 全日制普通高級中學教科書 (必修). 第1冊 (上) 1. 中國北京沙灘后街55號: 人民教育出版社. 2003: 75–79. ISBN 7-107-16755-3 （中文（中國大陸））.
2. 人民教育出版社中學數學室. 第2章「函數」第2.7節「對數」中的「閱讀材料」部分. 數學. 全日制普通高級中學教科書 (必修). 第1冊 (上) 1. 中國北京沙灘后街55號: 人民教育出版社. 2003: 80–81. ISBN 7-107-16755-3 （中文（中國大陸））.