对数

1 定义

若a^x = b,则我们设log_a^b = x,读做以a为底的b的对数为x,其中log称为对数公式,用来求取一个数a的多少次方为b

2 基本性质

显然,根据定义得到:

loga1=0(1) log_a^1 = 0 \tag1
logaa=1(2) log_a^a = 1 \tag2
logaax=x(3) log_a^{a^x} = x \tag3
logalogab=b(4) log_a^{log_a^b} = b \tag4

3 运算法则

logaM÷N=logaM−logaN(5)log_a^{M \div N} = log_a^M - log_a^N \tag5

Proof:

DefDef x=logaM,y=logaN,α=logaM⋅Nx=log_a^M,y = log_a^N,\alpha = log_a^{M \cdot N} thenthen:

  • ax=Ma^x = M
  • ay=Na^y = N
  • aα=M⋅Na^{\alpha} = M \cdot N
M÷N=ax÷ay=ax−y M \div N = a^x \div a^y = a^{x-y}

两边同时取对数

logaM÷N=logaax−y=x−y=logaM−logaN log_a^{M \div N} = log_a^{a^{x-y}} = x-y = log_a^M - log_a^N

Q.E.D


logaM⋅N=logaM+logaN(6)log_a^{M \cdot N} = log_a^M + log_a^N \tag6

DefDef x=logaM,y=logaN,α=logaM⋅Nx=log_a^M,y = log_a^N,\alpha = log_a^{M \cdot N} thenthen:

  • ax=Ma^x = M
  • ay=Na^y = N
  • aα=M⋅Na^{\alpha} = M \cdot N
M⋅N=ax⋅ay=ax+y M \cdot N = a^x \cdot a^y = a^{x+y}

Proof:

两边同时取对数

logaM⋅N=logaax+y=x+y=logaM+logaN log_a^{M \cdot N} = log_a^{a^{x+y}} = x+y = log_a^M + log_a^N

Q.E.D


logaMn=n⋅logaM(7)log_a^{M^n} = n \cdot log_a^M \tag7

Proof:

Def: x=logaMx = log_a^M

ax=M(ax)n=Mnax⋅n=Mn \begin{aligned} a^x &= M \\ (a^x)^n &= M ^n \\ a^{x\cdot n} &= M ^n \end{aligned}

两边现时取对数

logaMn=logaax⋅n=x⋅n=n⋅logaM log_a^{M ^n} = log_a ^{a^{x\cdot n} } = x \cdot n = n \cdot log_a^M

Q.E.D


loga1÷N=0−logaN=−logaN log_a{1 \div N} = 0 - log_a^N = -log_a^N

4 换底公式

log⁡ab=log⁡cblog⁡ca\log_{a}^b = \frac{\log_c^b}{\log_c^a}

证明:

设x = \log_a^b

x=log⁡ab→ax=bx = \log_a^b \rightarrow a^x = b

对a^x = b 式子的两边取对数,得到

logcax=logcb→x⋅logca=logcb\begin{matrix} log_c^{a^x} &=& log_c^b \\ &\rightarrow& x \cdot log_c^a = log_c^b \end{matrix}

两边同时除以log_c^a,得到

x=logcblogcalog⁡ab=logcblogca\begin{matrix} x &=& \frac{log_c^b}{log_c^a} \\ \log_a^b &=& \frac{log_c^b}{log_c^a} \end{matrix}

5 推论

TODO

6 题目

最多分解次数为⌈log2n⌉\lceil log_2^n \rceil

使用数学归纳法

7 相关题解

暂无相关题解