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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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  • $\begingroup$ Anti-log of $y\log x$. $\endgroup$ – Gerry Myerson May 16 '13 at 13:04
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    $\begingroup$ 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. en.wikipedia.org/wiki/Exponentiation_by_squaring $\endgroup$ – Thomas Andrews May 16 '13 at 13:10
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    $\begingroup$ @AlbertRenshaw, "exponentiation by squaring" involves no exponentiantion, just multiplications. $\endgroup$ – vonbrand May 16 '13 at 13:15
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    $\begingroup$ 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. $\endgroup$ – Gerry Myerson May 16 '13 at 13:15
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    $\begingroup$ @GerryMyerson, does information theory not count as grownup mathematics? It uses base 2 more than $e$. $\endgroup$ – Peter Taylor May 16 '13 at 14:14
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You can always use the Taylor series for $f(u) = e^u$.

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

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

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