What's the super logarithmic inverse of tetration for $\bf{^{2}{x}}$?

Is it $slog^{x}_{2}$?

  • $\begingroup$ I presume you've already looked into previous work by Ioannis Galidakis? $\endgroup$ – J. M. is a poor mathematician Apr 18 '12 at 8:31

The inverse function of ${^x{2}}$ is the base-2 super-log of x, which is what you wrote. The inverse function of ${^2{x}}$ is the 2nd super-root of x.

While the super-log seems to have a more commonly accepted notation (slog), there is no such notation for super-root (or, if you wish, dozens of notations). There are also some places (like Wikipedia's Tetration page) that call the 2nd super-root the "super-sqrt" function. But that's OK, because lots of things in mathematics don't have unique notations. Take $e$ for example. It is used for basis vectors, physical constants, dimensional units, other variables, and the "base of the natural logarithm".

So if you're looking for terminology, the inverse of ${^2{x}}$ is called "super-root", but if you're looking for notation, just use $f(x)$, along with the explanation: "where $f$ is the second super-root".

  • $\begingroup$ Thanks Andrew, I think that helped solve my problem. One question though: you mentioned that the inverse of $\bf{^2}{x}$ to be the 2$^{nd}$ super-root of $x$. Would my $\bf{slog^x_2}$ still satisfy? $\endgroup$ – A T Apr 18 '12 at 17:57
  • 1
    $\begingroup$ No, super-roots and super-logs are different functions, they do not satisfy each other's equations. $\endgroup$ – Andrew Robbins Apr 18 '12 at 18:00
  • $\begingroup$ Apologies, it's 4AM here. Only after rereading your answer did I realise you were talking about super roots on the same paragraph that you discussed super logarithms. $\endgroup$ – A T Apr 18 '12 at 18:01
  • $\begingroup$ +1 and accepted. Thanks for your help $\endgroup$ – A T Apr 18 '12 at 18:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.