proof involving log

[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

 

[pka]

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

 

[Guest]

what does "ln" mean? or what is it?..

 

[pka]

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.

 

[soroban]

Hello, bittersweet!

For \(\displaystyle a\,>\,0,\;a\,\neq\,0\), prove that: \(\displaystyle \log_a\left(\frac{1}{x}\right)\:=\:\log_{\frac{1}{a}}(x)\)

Let \(\displaystyle p\:=\:\log_a\left(\frac{1}{x}\right)\;\;\Rightarrow\;\;a^p\:=\:\frac{1}{x}\;\;\Rightarrow\;\;x\:=\:\frac{1}{a^p}\)

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

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

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