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This is kind of a spinoff on my question Divide by a number without dividing.

Can anyone think of some clever ways to raise any given number to any given power without using an exponent anywhere in your equation/formula?

$$x^{y}=z$$

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Albert Renshaw
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  • Anti-log of $y\log x$. – Gerry Myerson May 16 '13 at 13:04
  • One thing I was thinking of was $\int{y*x}$ but you would end up with a $±Constant$ – Albert Renshaw May 16 '13 at 13:04
  • @GerryMyerson Wouldn't it be the anti-log base x of y? Either way that one's a little bit sketchy, an anti-log is just exponents basically. But it still works so I'll plus one that. (User's must keep in mind that the exponent in the anti-log formula is just the notation for writing anti-log, you are not actually raising it to the negative first power. $log^{-1}_{x}(y)$ – Albert Renshaw May 16 '13 at 13:10
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    Your title says $n$th power, which implies $n$ is an integer, but your question says $x^y$ where $y$, by implication, is not an integer. Integer powers can be efficiently compute using the Exponentiation by Squaring method. http://en.wikipedia.org/wiki/Exponentiation_by_squaring – Thomas Andrews May 16 '13 at 13:10
  • @ThomasAndrews well I meant for all numbers, not just integers, I thought that a double stricken "z" stood for integers only, not an "n", anyways, the exponentiation by squaring still involves exponents in the formula. – Albert Renshaw May 16 '13 at 13:12
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    @AlbertRenshaw, "exponentiation by squaring" involves no exponentiantion, just multiplications. – vonbrand May 16 '13 at 13:15
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    There's only one base in grownup mathematics, and that's $e$. Anti-log of $Q$ is a way of writing $e^Q$ without writing an exponent. – Gerry Myerson May 16 '13 at 13:15
  • The term $n$th is an ordinal, which implies that $n$ is a natural number. We never say "I finished in $\sqrt{2}$th place in the race!" Also, while $\mathbb Z$ represents the set of integers, $n$ as a variable tends to imply a natural number or integer. Not a hard rule, just a convention. But the real problem is that you switch variables and never specify what you mean in either case. – Thomas Andrews May 16 '13 at 13:16
  • @ThomasAndrews Corrected the title! Thanks for your help :o) – Albert Renshaw May 16 '13 at 14:10
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    @GerryMyerson, does information theory not count as grownup mathematics? It uses base 2 more than $e$. – Peter Taylor May 16 '13 at 14:14
  • @Peter, OK, I'll give you base 2. – Gerry Myerson May 17 '13 at 09:53

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You can always use the Taylor series for $f(u) = e^u$.

$$ x^y = 1 + y \ln x + \frac{(y \ln x)(y \ln x)}{2!} + \frac{(y \ln x)(y \ln x)(y \ln x)}{3!} + \cdots $$

Caleb Stanford
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  • Very nice! :) Factorials have always fascinated me, it seems like they would be an ideal way to exponentiate since they involve sets of multiplication. – Albert Renshaw May 16 '13 at 14:11
  • Now that I understand more intuitively what logarithms actually are I redact my above comment. This is still a nice solution though :D – Albert Renshaw Oct 26 '15 at 20:29