<> product , merchant , Logarithm of power

logaMN=logaM+logaNlog_{a}MN=log_{a}M + log_{a}Nloga​MN=loga​M+loga​N The derivation process is as follows .

prove ： set up logaM=p,logaN=q be ap=M,aq=N, Substitution logaMN, have to logaMN=loga(ap⋅aq)=logaap+q=p+q=logaM+l
ogaN therefore ：logaMN=logaM+logaN \begin{array}{ll} prove ： set up log_{a}M=p,\quad log_{a}N=q \\
log_{a}MN=log_{a}(a^p\cdot a^q) = log_{a}a^{p+q} = p+q = log_{a}M + log_{a}N \\
therefore ：log_{a}MN=log_{a}M + log_{a}N \end{array} prove ： set up loga​M=p,loga​N=q be ap=M,aq=N, Substitution lo
ga​MN, have to loga​MN=loga​(ap⋅aq)=loga​ap+q=p+q=loga​M+loga​N therefore ：loga​MN=loga​M+loga​N​

logaMN=logaM−logaNlog_{a}{\Large\frac{M}{N}}= log_{a}M - log_{a}Nloga​NM​=loga​
M−loga​N The derivation process is as follows .

prove ： set up logaM=p,logaN=q be ap=M,aq=N, Substitution logaMN, have to logaMN=loga(apaq)=logaap−q=p−q=logaM−lo
gaN therefore ：logaMN=logaM−logaN \begin{array}{ll} prove ： set up log_{a}M=p,\quad log_{a}N=q \\
log_{a}{\frac{M}{N}}=log_{a}(\frac{a^p}{a^q}) = log_{a}a^{p-q} = p-q = log_{a}M
- log_{a}N \\ therefore ：log_{a}\frac{M}{N}=log_{a}M - log_{a}N \end{array} prove ： set up loga​M=p,
loga​N=q be ap=M,aq=N, Substitution loga​NM​, have to loga​NM​=loga​(aqap​)=loga​ap−q=p−q=loga​M−loga​N
therefore ：loga​NM​=loga​M−loga​N​

logaMb=b⋅logaMlog_aM^b=b\cdot log_aMloga​Mb=b⋅loga​M The derivation process of

prove ： set up logaM=p be ap=M, Substitution logaMb have to logaMb=loga(ap)b=logaapb=pb=b⋅logaM
\begin{array}{ll} prove ： set up log_aM=p \\ be a^p=M ,\quad Substitution log_aM^b \\
have to log_aM^b=log_a(a^p)^b = log_aa^{pb} = pb = b \cdot log_aM \end{array} prove ： set up loga​M
=p be ap=M, Substitution loga​Mb have to loga​Mb=loga​(ap)b=loga​apb=pb=b⋅loga​M​

alogaM=Ma^{log_aM} = Maloga​M=M The derivation process of

prove ： set up logaM=p be ap=M, Substitution alogaM have to alogaM=alogaap=ap=M \begin{array}{ll} prove ： set up log_aM=p
\\ be a^p=M ,\quad Substitution a^{log_aM} \\ have to a^{log_aM} = a^{log_aa^p} = a^p = M
\end{array} prove ： set up loga​M=p be ap=M, Substitution aloga​M have to aloga​M=aloga​ap=ap=M​

<> Bottom changing formula

logbN=logaNlogablog_bN=\Large\frac{log_aN}{log_ab}logb​N=loga​bloga​N​ The derivation process is as follows .
prove ： set up logbN=x, be bx=N Take both sides at the same time a Base logarithm logabx=logaNx⋅logab=logaNx=logaNlogab therefore logbN=logaN
logab \begin{array}{ll} prove ： set up log_bN=x, be b^x = N \\ Take both sides at the same time a Base logarithm \\ log_a{b^x} =
log_aN \\ x\cdot log_ab = log_aN \\ x = \frac{log_aN}{log_ab} \\
therefore log_bN=\frac{log_aN}{log_ab} \end{array} prove ： set up logb​N=x, be bx=N Take both sides at the same time a Base logarithm loga​bx=
loga​Nx⋅loga​b=loga​Nx=loga​bloga​N​ therefore logb​N=loga​bloga​N​​

<> Other formulas

log⁡ana=1n\displaystyle \log_{a^n}a = \frac{1}{n}logan​a=n1​ The derivation process of

prove ： set up logana=p be (an)p=a Namely anp=a Take constant logarithm on both sides at the same time ,lganp=lganp⋅lga=lga,np=1,p=1n therefore logana=1n
\begin{array}{ll} prove ： set up log_{a^n}a = p \\ be \quad (a^n)^p = a Namely a^{np} = a \\
Take constant logarithm on both sides at the same time ,\quad lg{a^{np}}=lg{a} \\ np\cdot lga = lga, \quad np = 1,\quad p =
\frac{1}{n} \\ therefore log_{a^n}a = \frac{1}{n} \end{array} prove ： set up logan​a=p be (an)p=a Namely anp=a
Take constant logarithm on both sides at the same time ,lganp=lganp⋅lga=lga,np=1,p=n1​ therefore logan​a=n1​​

1logab=logba{\Large \frac{1}{log_ab}} = log_baloga​b1​=logb​a The derivation process of .
Here we use the bottom changing formula , The process is relatively simple

1logab=1lgblga=lgalgb=logba \frac{1}{log_ab} =
\frac{1}{\frac{lgb}{lga}}=\frac{lga}{lgb}=log_baloga​b1​=lgalgb​1​=lgblga​=logb​
a

loganM=1n⋅logaMlog_{a^n}M = {\Large \frac{1}{n}} \cdot log_aMlogan​M=n1​⋅loga​M
The derivation process of

prove ： set up logaM=p be ap=M, Substitution loganMloganap=p⋅logana=1n⋅p=1n⋅logaM prove ： set up log_aM = p \\ be a^p
= M, Substitution log_{a^n}M \\ log_{a^n}a^p=p \cdot log_{a^n}a = \frac{1}{n} \cdot p =
\frac{1}{n} \cdot log_aM prove ： set up loga​M=p be ap=M, Substitution logan​Mlogan​ap=p⋅logan​a=n1​⋅p=n1​⋅
loga​M

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