Is this a property of log :

$$\log_{1/x^3} x = -\frac{1}{3} \log_{x} x$$

I was looking through the solution of a sum I was solving the the only way the steps made sense was if this took place. I am confused because I have never come across properties of the base of a logarithm either online or in any of my reference books. Please help me out.

This is the part of the solution where I think the above property was applied.

$$2^{-\log_{1/8}124} =2^{\log_{2}5}\, 2^{-1}$$

First $2^{-1}$ was separated and then for the remaining it was written as a cube and then $3$ was taken out and cancelled. Does such a transformation exist?

  • 1
    $\begingroup$ Recall that $\log_ab=c$ if and only if $a=b^c$. Thus, $\log_{1/x^3}x=a$ if and only if $\left(\frac1{x^3}\right)^a=x$. Now $\left(\frac1{x^3}\right)^{-1}=x^3$, so $$x=(x^3)^{1/3}=\left(\left(\frac1{x^3}\right)^{-1}\right)^{1/3}=\left(\frac1{x^3}\right)^{-1/3}\;,$$ and it is indeed true that $\log_{1/x^3}x=-\frac13$. (And $\log_xx=1$ for any $x$ for which logarithms to the base $x$ make sense.) This is really just the change of base formula. $\endgroup$ Apr 30 '20 at 17:28

Just use the base change for logarithm: you have in general that $ \log_a b=\frac{\log_c b}{\log_c a}$, so $$\log_{1/x^3} x =\frac{\log_x x}{\log_{x} \frac{1}{x^3}}=-\frac{1}{3} \log_x x$$ Having used the other properties: $\log u^v = v \log u$ and $\log \frac{1}{w}=-\log w$.


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