Is $\log^{-1}n = \frac{1}{\log n}$? I've encountered a solution to a problem where the above step was used. I can't determine the validity of the above line. I have thought a lot about it, but can't prove whether it is right or wrong. Please help?
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$\begingroup$ Take $n=e$, then if it was the inverse function you would have $e^e = 1$. $\endgroup$– copper.hatNov 4, 2016 at 5:36
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$\begingroup$ You could edit the question to show more context, such as the step that included $\log^{-1} n$ and the step before it. The notation seems unusual, and maybe so is the interpretation of the notation in that "solution". $\endgroup$– David KNov 4, 2016 at 5:43
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2 Answers
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Never $\log^{-1} n$ means $\text{antilogarithm}(n) $ i.e $e^n$ because $\log(e^n)=n$ and $\frac{1}{n}$ means inverse of $n$ .
So they are not the same
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$\begingroup$ If we'd need to remove the denominator, what could we rewrite $$\frac{1}{\log n}$$ as? I want an expression which has a numerator only. Is it possible? $\endgroup$ Nov 4, 2016 at 5:41
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1$\begingroup$ If you really must, then you can use a change of base formula to say that $\log^{-1}_e (n) = \log_n (e)$. $\endgroup$– ConManNov 4, 2016 at 5:42
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$\begingroup$ I am not getting it ;can you please rewrite using $$ $\endgroup$ Nov 4, 2016 at 5:42
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1$\begingroup$ @SomenathSinha Some people might write $\log\log n$ as $\log^2 n$ but I don't think I've ever seen it. I think a notation like that would normally be explained in the same document before it was used, in order to avoid misunderstanding. $\endgroup$– David KNov 4, 2016 at 5:50
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Ouch!
This is very classic ambiguous notation. The notation $f^{-1}(x)$ can mean one of two entirely different meanings depending on context.
It could (but usually doesn't) mean $(f(x))^{-1} = \frac 1{f(x)}$. The is the reciprical of $f$. It is simply the function as a value raised to the $-1$ power.
More frequently it means the functional inverse (if there is one). $f^{-1}(x) = y$ means that $y$ is the value so that $f(y) = x$. For example, if $f(x) = x^3 + 2$ then $f^{-1}(x) = \sqrt[3]{x - 2}$ because if $y = \sqrt[3]{x - 2}$ then $f(y) = x$. And $f(f^{-1}(x)) =x $ by definition.
These two interpretations have NOTHING to do with each other and if you do confuse them (as people often do) you get utterly confusing contradictory nonsense. Example if $f(x) = x^3 + 2$ and we thought both $f^{-1}(x) = \frac 1{f(x)}=\frac 1{x^3 + 2}$ and that $f^{-1}(y) = \sqrt[3]{x-2}$ and THAT MAKES NO FREAKING SENSE WHATSOEVER. (You will not BELIEVE the number of students who get frustrated because that makes no sense at all to them, but are convinced it must be so and it must be their fault that they simply can't understand it.)
So if you are interpreting $\log^{-1} x$ as $\frac 1{\log x}$ (and I highly suggest you do NOT do so-- if you want notation for $\frac 1{\log x}$ use $(\log x)^{-1}$ inside parenthesis instead) then, the answer to the question as ... yeah.
But if you are interpreting $\log^{-1} x$ as the inverse function--- that is, the function so that if $y = \log x$ then $\log^{-1} y = \log^{-1} \log x = x$ (and this is how you SHOULD interpret it) then the answer is an emphatic !!!!NO!!!!!!. As it turns out, $\log^{-1} y = e^y$ (assuming $\log$ is the natural log, base $e$-- if $\log $ it that base 10 log then $\log_{10}^{-1} y = 10^y$).
To confuse the two and think $\log^{-1}(\log y) = y$ and $\log^{-1}(\log y) = \frac 1{\log y} = y$ will result in meaningless, impossible and unresolvable GARBAGE. So don't ever confuse the two.
And I STRONGLY urge you to NEVER use $f^{-1}(x)$ to mean the reciprical but to ALWAYS mean the functional inverse.
So $\log_e^{-1} y = e^y$ and $\log_e^{-1} y \ne \frac 1{\log_e y}$.
