对数性质的证明 – 源码巴士

For all real numbers a>0, b>0, c>0,M>0, N>0

$log_aM + log_aN = log_aMN$ (1)

proof:

assume: $log_aM = m, log_aN = n$

so: $a^m = M, a^n = N \Rightarrow M \cdot N = a^m \cdot a^n = a^{m+n}$

$log_a{MN} = m + n = log_aM + log_aN$

$log_aM - log_aN = log_a\frac{M}{N}$ (2)

proof:

assume: $log_aM = m, log_aN = n$

so: $a^m = M, a^n = N \Rightarrow \frac{M}{N}=\frac{a^m}{a^n} = a^{m - n} \Rightarrow log_a{\frac{M}{N}} = m - n = log_aM - log_aN$

$log_aa^M = M$ (3)

proof:

assume: $a^M = B \Rightarrow log_aB = M$

so: $log_aa^M = M$

$a^{log_aM} = M$ (4)

proof:

assume: $log_aM = B$

so: $a^B = M$

$\because B = log_aM, a^B = M$

$\therefore a^B = a^{log_aM} = M$

$log_aM^N = Nlog_aM$ (5)

proof:

$log_aM^N = log_a(\overbrace{M \cdot M \cdots M}^{N})$

$\because log_aM + log_aN = log_aMN$ property (1)

$\therefore log_a(\overbrace{M \cdot M \cdots M}^{N}) = \overbrace{log_aM + log_aM + \cdots log_aM}^{N} = Nlog_aM$

$log_ab = \frac{log_cb}{log_ca}= \frac{lnb}{lna} = \frac{lgb}{lga}$ (6)

proof:

assume: $log_ab = m$ then $b=a^m, log_cb = log_ca^m, log_ca^m = mlog_ca$ property (5)

$\because log_cb = mlog_ca => \frac{log_cb}{log_ca} = m$ and $m = log_ab$

$\therefore log_ab = \frac{log_cb}{log_ca}$

$log_ab = \frac{lnb}{lna}$ means c = e

$log_ab = \frac{lgb}{lga}$ means c = 10

$log_{a^M}b^N = \frac{N}{M}log_ab$ (7)

proof:

$\because log_{a^M}{b^N} = \frac{log_cb^N}{log_ca^M}$ property (6)

$\frac{log_cb^N}{log_ca^M} = \frac{Nlog_cb}{Mlog_ca}$ property (5)

$\frac{Nlog_cb}{Mlog_ca} = \frac{N}{M} \frac{log_cb}{log_ca}$

$\frac{N}{M} \frac{log_cb}{log_ca} = \frac{N}{M}log_ab$ property (6)

$\therefore log_{a^M}{b^N} = \frac{N}{M}log_ab$

$log_{\frac{1}{a}}b = log_a{\frac{1}{b}}$ (8)

$\because log_{\frac{1}{a}}b = log_{a^{-1}}b = log_{a^{-1}}b^1 = \frac{1}{-1}log_ab$ property (7)

$\frac{1}{-1}log_ab = -1log_ab = \frac{-1}{1}log_ab = log_{a^1}{b^{-1}} = log_a{\frac{1}{b}}$

$\therefore log_{\frac{1}{a}}b = log_a{\frac{1}{b}}$

$log_ba = \frac{1}{log_ab}$ (9)

proof:

assume: $x = log_ab, y = log_ba$

$\Rightarrow a^x = b, b^y = a$

$log_ab = log_a{a\cdot\frac{b}{a}} = log_aa + log_a{\frac{b}{a}}$ property (1)

$log_aa + log_a{\frac{a^x}{b^y}} = log_aa + log_a{a^x} - log_a{b^y}$ property (2)

$log_aa + xlog_aa - ylog_ab$ property (5)

$1+x - ylog_ab$

$ylog_ab \Rightarrow ylog_ab = 1$

$\because log_ab = x$

$\therefore yx = 1 \Rightarrow y = \frac{1}{x}$

$\because x = log_ab, y = log_ba$

$\therefore log_ba = \frac{1}{log_ab}$

$a^{log_bc} = c^{log_ba}$ (10)

proof:

Take $log_b$ on both sides:

$a^{log_bc} = c^{log_ba}$

$\Longleftrightarrow log_b{a^{log_bc}} = log_b{c^{log_ba}}$

$\Longleftrightarrow log_bc \cdot log_ba = log_ba \cdot log_bc$ property (5)

$\because log_bc \cdot log_ba = log_ba \cdot log_bc$

$\therefore a^{log_bc} = c^{log_ba}$

原文链接：https://blog.csdn.net/jun2016425/article/details/125230274