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Find maximum root of the equation $$x - \frac{1000}{\log 2} \log x = 0$$ It locates between $13746$ and $13747$, but I want to find right solution not using graphing calculators. Thanks in advance.

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  • $\begingroup$ dont think possible using elementary algebra $\endgroup$ – Archis Welankar Feb 7 '16 at 16:26
  • $\begingroup$ i think taylor series needed $\endgroup$ – Archis Welankar Feb 7 '16 at 16:26
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    $\begingroup$ @ArchisWelankar If a numerical approximation is all that's wanted, there are many ways to do it with just trial and error and a non-graphing calculator. $\endgroup$ – Zubin Mukerjee Feb 7 '16 at 16:30
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Consider function $$f(x)=x - \frac{1000}{\log 2} \log (x)$$ $$f'(x)=1-\frac{1000}{x \log (2)}$$ The function goes through a minimum (by second derivative test) at $x=\frac{1000}{\log (2)}$.

So, let us start Newton method which will generate the following iterates $$x_0=2000$$ $$x_1=34175.5$$ $$x_2=14218.2$$ $$x_3=13747.7$$ $$x_4=13746.8$$ which is the solution for six significant figures.

Edit

What I tried to show above is that, even with a very poor estimate of the solution, we can get the solution in few iterations of Newton method. However, we can improve the process since, for large values of $k$, the root of equation $f(x)=x-k\log(x)=0$ can be estimated as $$x\approx k\Big(\log(k)+\log\big(\log(k)\big)\Big)$$ For $k= \frac{1000}{\log 2}$, this will give $x_0= 13357.4$ and the solution would have been obtained after two iterations.

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Solve $$x-\frac{1000}{\log(2)}\log(x)=0$$ for $x$. Substitute $x=e^t$: $$e^t-\frac{1000}{\log(2)}t=0;$$ subtract $e^t$ from both sides: $$-\frac{1000}{\log(2)}t=-e^t;$$ multiply both sides by $\dfrac{\log(2)}{1000}$: $$-t=-\frac{\log(2)}{1000}e^t;$$ divide both sides by $e^t$: $$-t\frac{1}{e^t}=-\frac{\log(2)}{1000};$$ rewrite $1/e^t=e^{-t}$: $$-te^{-t}=-\frac{\log(2)}{1000};$$ substitute $u=-t$: $$ue^u=-\frac{\log(2)}{1000};$$ take the branch $-1$ product log of both sides: $$u=\operatorname{W}_{-1}\left(-\frac{\log(2)}{1000}\right);$$ substitute back for $u$: $$-t=\operatorname{W}_{-1}\left(-\frac{\log(2)}{1000}\right);$$ multiply both sides by $-1$: $$t=-\operatorname{W}_{-1}\left(-\frac{\log(2)}{1000}\right);$$ substitute back for $t$: $$\log(x)=-\operatorname{W}_{-1}\left(-\frac{\log(2)}{1000}\right);$$ take exponentials of both sides: $$x=e^{-\operatorname{W}_{-1}\left(-\log(2)/1000\right)}\phantom{.};$$ write the answer out: $$\therefore x=\boxed{e^{-\operatorname{W}_{-1}\left(-\log(2)/1000\right)}\phantom{.}}\approx13746.809166647028808721383407435.$$

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  • $\begingroup$ The branch $-1$ was used because that was the root desired by the OP. Notice that the branch $0$ also satisfies the equation, but none of the others do courtesy of the penultimate equation given $\forall_{z\in\mathbb{C}}\Im\left(\log\left(z\right)\right)\in{({{-\pi};{\pi}}]}$. For conventions adopting other branch cuts of $\log$, other solutions may be conveyed. $\endgroup$ – dbanet Feb 7 '16 at 17:01
  • $\begingroup$ For more information considering the product log function $\operatorname{W}$, refer to Wolfram MathWorld and Wikipedia. $\endgroup$ – dbanet Feb 7 '16 at 17:10
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$$x-\frac{1000\ln(x)}{\ln(2)}=0\Longleftrightarrow$$ $$-\frac{1000\ln(x)}{\ln(2)}=-x\Longleftrightarrow$$ $$\frac{1000\ln(x)}{\ln(2)}=x\Longleftrightarrow$$ $$1000\ln(x)=x\ln(2)\Longleftrightarrow$$ $$e^{1000\ln(x)}=e^{\ln(2)x}\Longleftrightarrow$$ $$x^{1000}=2^x\Longleftrightarrow$$ $$x=\exp\left[-\text{W}\left(-\frac{\ln(2)}{1000}\right)\right]\approx13746.809166647028809$$

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