$a \log_bx=\log_{b^\frac{1}{a}}x$

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$a \log_bx=\log_{b^\frac{1}{a}}x$

Is this a valid property of logarithms? Using other exponential and logarithmics laws can you explain why or why not

Logarithms

Bendollak

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This is not a valid property for logarithms. In general we have
\[a \log _b x= \log_b x^{\frac{1}{a}},\]
but your stated identity does not hold. For a counter example let $x=2, b=2, a=2$. Then
\[a \log _b x = 2 \log_2 2=2.\]
However
\[\log ^{\frac{1}{a}}_b x=(\log_b x )^{^{\frac{1}{a}}}= (\log_2 2 )^{^{\frac{1}{2}}}=1^{{\frac{1}{2}}}=1.\]
S
\[a \log _b x=2\neq 1= \log ^{\frac{1}{a}}_b x.\]

*****************
If you meant
\[a \log _b x= \log_{b^{\frac{1}{a}}}x (*)\]
then it is true. To show this, you can simply use the definition of logarithms. Let
\[A= a\log _b x \Rightarrow \frac{A}{a}= \log _b x \Rightarrow x= b^{\frac{A}{a}}\]
\[\Rightarrow x=(b^{\frac{1}{a}})^A \Rightarrow A= \log_{b^{\frac{1}{a}}}x.\]
Hence (*) holds.

Erdos

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Bendollak

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I think this is my fault. I meant to write the base alone is raised to the power of 1/a, not log_b in the second equation.
- Erdos
  
  +1
  
  Yes! I revised my solution. The statement is true. Let me know if you need any clarifications.
- Bendollak
  
  0
  
  Thank you Erdos that explains it perfectly
Walter

+1

I edited the title accordingly.

$a \log_bx=\log_{b^\frac{1}{a}}x$

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