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I am working through an algorithms workbook and I have the following equation: $$N^{\frac{e}{\sqrt{\log(N)}}}$$ I know I can simplify it somehow using the properties of logs and exponents but am a little rusty on that basic math. If you could provide explanation of your process I would appreciate it.

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Use the fact that $N=e^{\log{N}} $. Then

$$N^{\frac{e}{\sqrt{\log{N}}}} = e^{e \sqrt{\log{N}}}$$

Not much more I can think of doing here.

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  • $\begingroup$ Dang! That was a pretty bad error on my part this morning. Sheesh (I can't do arithmatic on a keyboard worth a damn!) $\endgroup$
    – fleablood
    Oct 26 '15 at 19:24
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$N^{\frac{e}{\sqrt{\log(N)}}}$

Take the log. $\log N^{\frac{e}{\sqrt{\log(N)}}} = \frac{e}{\sqrt{\log(N)}}\log(N) = e \sqrt{log(N)}$. Raise back to the exponent to get $N^{\frac{e}{\sqrt{\log(N)}}} = e^{ e \sqrt{log(N)}} $ which isn't much better but it has only a single instance of a variable. If we need to solve for N it will be easy.

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  • $\begingroup$ $a/\sqrt a \ne 1/\sqrt a$ $\endgroup$ Oct 26 '15 at 17:31
  • $\begingroup$ Dang!!!!! I know better than that! $\endgroup$
    – fleablood
    Oct 26 '15 at 19:22
  • $\begingroup$ It's those dang nested LaTex clauses. I somehow flipped the terms upside down. Stupid thing is once I made the error I couldn't see t to proof it. Very bad error on my part. $\endgroup$
    – fleablood
    Oct 26 '15 at 19:26

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