I was solving some exercises and the most of them I found the algebraic properties on the web.
But in the equation $n\log_2 n = 10^6$ I have no idea, I tried several ways to solve the equation and none of them worked.
Thanks!
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Welcome to Math.SE! There is no “algebraic formula” for this kind of equation. – egreg Mar 21 '14 at 00:09
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2$$ n\log_2n=\log_2n^n=10^6 $$ Thus, $$ n^n=2^{10^6} $$Hopefully this helps. – Mar 21 '14 at 00:09
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The solution of this kind of equation has no closed form. But it can be represented by a sort of special function, called Lambert W function. Write original equation as $$m2^m=10^6$$ by setting $m=2^n$, then the solution is $$m=W_2(10^6)$$ Or, $$n=\log_2(W_2(10^6))=\frac{10^6\ln2}{W_e(10^6\ln2)}$$
Shuchang
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Rearrange: $\log_2{n^n} = 10^6$
$n^n = 2^{10^{6}} = 2^{1000000}$
This equation can't be solved with elementary functions, you need to use the Lambert W function.
Approximating: $n \approx 62746.126$
qwr
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1Looks good and compares well with: 62,746.1264697 using Solver in MS Excel – Gary's Student Mar 21 '14 at 00:16
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Thank you very much! I even got to $n^n = 2^{10^6}$ but had no idea how to go beyond. And that's because there is no way without using the function :D. – user136913 Mar 21 '14 at 00:22
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As said in previous answers and comments, equations such as $$n\log_a n = b$$ have no elemental solutions. Only Lambert function provides a solution which is (for the algebraic case) $$n=\frac{b \log (a)}{W(b \log (a))}$$ For large values of $x$, $$W(x) \simeq \log (x)-\log (\log (x))+\frac{\log (\log (x))}{\log (x)}$$ So, in your case where $x=10^6 \log(2)$, this would give an estimate equal to $62766.1$ while the exact solution is $62746.1$.
By the way, any equation of the form $$a+b x+c \log (d x+e)=0$$ has solutions which can be expressed using Lambert function $$x=\frac{c d W\left(\frac{b e^{\frac{b e}{c d}-\frac{a}{c}}}{c d}\right)-b e}{b d}$$
Claude Leibovici
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