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I need to solve $\log n = \frac{\log 2}{10} \sqrt{n}$.

I know it is a transcendental function and also hear about generalizes Lambert function (Lambert W-function) could help me to solve it. But I have no idea about how to apply it.

Thomas Andrews
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Daniel Ferreira Castro
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4 Answers4

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Rewriting it as $$\left(\sqrt{n}\right)^{20} = 2^{\sqrt{n}}$$

Then let $m=\sqrt{n}$ and we need to solve:

$$m^{1/m} = 2^{1/20}$$

Taking the log of both sides, and letting $y=\log m=\frac{\log n}{2}$, you get:

$$ye^{-y} = \frac{\log 2}{20}$$

So $y = -W\left(-\frac{\log 2}{20}\right)$ and $n=e^{2y}$.

Thomas Andrews
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  • I missed some steps from $m=\sqrt{n}$ to $m^{1/m}=2^{1/20}$ – Daniel Ferreira Castro Aug 11 '14 at 21:33
  • $m^{20} = 2^m$. Now raise both sides to the power of $\dfrac 1{20m}$. – Thomas Andrews Aug 11 '14 at 21:34
  • Cool. I followed until the definition of $W=ye^y$. :) Almost there. Only missing the steps from $ye^y$ to the last line and how do I evaluate the function in order to find $y$ and then find $n$.

    I am trying to understand how Wolfram alpha evaluate this same function. This is to solve a problem of big-O notation for a function.

    One of the solutions is about n=20519.8 that is $\dfrac{400W_{-1}\left(-\dfrac{\log(2)}{20}\right)^2}{\log^2(2)}$ and this is what I am trying to find out how to get there.

     – Daniel Ferreira Castro Aug 12 '14 at 00:20
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$10\log{n}=\log{(2\sqrt{n})}$

$\log{n^{10}}=\log{(2\sqrt{n})}$

$n^{10}=2\sqrt{n}$

If this is the right interpretation of your question

Paul Sundheim
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  • THe correct interpretation of my question is $n^{10} = 2^{\sqrt{n}}$. :)

    This is what I am trying to solve :)

     – Daniel Ferreira Castro Aug 11 '14 at 16:40
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    @DanielFerreiraCastro You could have answered my question and avoided a lot of people wasting their time by reformatting your question to be clearer. – Thomas Andrews Aug 11 '14 at 16:42
  • Paul, your sentence was "if this is the right interpretation..." Than I just replied saying what I was trying to find out. I thank you to point that to me because if it led to a wrong interpretation than I need to reformulate my text for a better understanding. And since text has no voice tone a simple sentence can be read in diferent ways. One way is to use the smilley to try to make the sentence a little more mild. I do not have any intention of offend no one, please accept my appologies if for any trouble I may have cause you. It was not my intention. – Daniel Ferreira Castro Aug 11 '14 at 16:48
  • And no offence was intended nor taken. It's all good :) – Paul Sundheim Aug 11 '14 at 16:51
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We have $$\log (n) = \dfrac{1}{10} \log(2 \sqrt{n}) \implies 10 \log(n) = \log(n^{10}) =\log(2 \sqrt{n}), $$ now since $\log $ is injective this implies that $$n^{10} = 2 \sqrt{n} = 2n^{1/2} \implies n^{10-1/2} = n^{19/2} = 2 \implies n = 2^{2/19}.$$

Surb
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Paul and Surb already answered if you meant $\log (2\sqrt n)$ but if you meant $\log_2\sqrt n$ instead, here's the solution.

$$\log n=\frac1{10}\log_2\sqrt{n}$$ Since $\log_a b=\dfrac{\log b}{\log a}$ $$\log n=\frac{\log\sqrt{n}}{10\log 2}$$ $$\log n-\frac{\log\sqrt{n}}{10\log 2}=0$$ Bring together the terms in one fraction. $$\frac{10\log 2\log n - \log\sqrt{n}}{10\log 2}=0$$ Multiply both sides by $10\log 2$ $$10\log 2\log n - \log\sqrt{n}=0$$ Since $n\log a=\log a^n$ $$\log n^{10\log 2} - \log\sqrt{n}=0$$ Since $\log a-\log b=\log \dfrac ab$ $$\log \left(\dfrac{n^{10\log 2}}{\sqrt n}\right)=0$$ Since $\dfrac{a^b}{a^c}=a^{b-c}$ and $\sqrt n=n^\frac 12$ $$\log \left(n^{10\log 2-\frac12}\right)=0$$ Take $\exp$ of both sides $$n^{10\log 2-\frac12}=1$$ Raise both side to the power $\dfrac{1}{10\log 2-\frac12}$
Note that $1^a=1$ for $a\ne0$ $$n=1$$

Alice Ryhl
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