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Supposedly, you can calculate the $n^{\text{th}}$ root of any number using logarithms. Where you find the log of the number under the radicand, divide that by $n$ and take the antilog of that.

I tried that with $\sqrt[5]{37}$, and something went wrong because I got an output of approximately $5$ even though Wolfram Alpha says it's about $2.0589\ldots$.

The code I entered into WolframAlpha was AntiLog(Log(37)/5).

Question: What went wrong, and what is the correct way to do this?

EDIT: If possible, could you also include a website or pdf or something where you can look up $\log(x)$ and $\text{antilog}(x)$.

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  • $\begingroup$ Unfortunately, WolframAlpha uses base $e$ for AntiLog, but base $10$ for Log. $\endgroup$ – Steve Kass Nov 27 '16 at 2:46
  • $\begingroup$ Typing "exp(ln(37)/5)" will give you what you want. $\endgroup$ – kccu Nov 27 '16 at 2:49
  • $\begingroup$ I have edited the question to include an extra request. Hope you don't mind! $\endgroup$ – Frank Nov 27 '16 at 2:52
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Wolfram Alpha says it's "Assuming "Log" is the natural logarithm." If you try $e^{\log(37)/5}$, you'll get what you're expecting.

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