proof involving log

G

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Guest
For a>0, a can not equal 1, x>0, prove that logsubscript a(1/x)= log subscript (1/a) X

ya..whaaaa..

thanks for the help in advance
 
These are true: \(\displaystyle \L
\begin{array}{l}
\log _a \left( {\frac{1}{x}} \right) = \frac{{\ln \left( {\frac{1}{x}} \right)}}{{\ln (a)}} = \frac{{ - \ln (x)}}{{\ln (a)}} \\
\log _{\frac{1}{a}} \left( x \right) = \frac{{\ln (x)}}{{\ln \left( {\frac{1}{a}} \right)}} \\
\end{array}\)

You finish this off
 
bittersweet said:
what does "ln" mean? or what is it?..
That is the natural logarithm.
It happens to be the only logarithm many of us think is worth teaching.
 
Hello, bittersweet!

For a>0,  a0\displaystyle a\,>\,0,\;a\,\neq\,0, prove that: loga(1x)=log1a(x)\displaystyle \log_a\left(\frac{1}{x}\right)\:=\:\log_{\frac{1}{a}}(x)
Let p=loga(1x)        ap=1x        x=1ap\displaystyle p\:=\:\log_a\left(\frac{1}{x}\right)\;\;\Rightarrow\;\;a^p\:=\:\frac{1}{x}\;\;\Rightarrow\;\;x\:=\:\frac{1}{a^p}

Let q=log1a(x)        (1a)q=x        x=1aq\displaystyle q\:=\:\log_{\frac{1}{a}}(x)\;\;\Rightarrow\;\;\left(\frac{1}{a}\right)^q\:=\:x\;\;\Rightarrow\;\;x\:=\:\frac{1}{a^q}

Hence: 1ap=1aq        ap=aq        p=q\displaystyle \,\frac{1}{a^p}\:=\:\frac{1}{a^q}\;\;\Rightarrow\;\;a^p\:=\:a^q\;\;\Rightarrow\;\;p\:=\:q

Therefore: loga(1x)  =  log1a(x)\displaystyle \log_a\left(\frac{1}{x}\right)\;=\;\log_{\frac{1}{a}}(x)
 
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