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I'm trying to find a solution for the above equation but can't find a good way short of trying to expand $ \log x$.

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  • $\begingroup$ That looks to me like an exact value would require "Lambert's w function" (defined as the inverse function to $xe^x$). There will not be any elementary way to solve it. $\endgroup$ – user247327 Sep 10 '18 at 0:18
  • $\begingroup$ I want to express x in terms of c. So I don't need an exact value. $\endgroup$ – caffiend Sep 10 '18 at 0:28
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Given the tags, I would expect that the most you actually need is to show that for every constant $c$ the inequality holds for all sufficiently large $x$. For asymptotics it doesn't matter exactly when it starts holding.

To do this, you might decide from the outset that "sufficiently large" is always going to be at least $1000$, in which case $$ x\log x + x - 1000 \ge x\log x $$

Then a sufficient condition to get what you want is that $x\ge 1000$ and additionally $$ x\log x > xc $$

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  • $\begingroup$ Thank you! This was the correct direction for me. $\endgroup$ – caffiend Sep 10 '18 at 2:39
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Consider the function $$f(x)=x\log (x) + x - 1000 -c x $$ $$f'(x)=\log (x)+2-c$$ $$f''(x)=\frac 1 x > 0 \qquad \forall x$$

The first derivative cancels for $$x_*=e^{c-2}\implies f(x_*)=-e^{c-2}-1000 <0$$ So, $x_*$ corresponds to a minimum.

The point where $f(x)=0$ is given in terms of Lambert function $$x_{sol}=\frac{1000}{W\left(1000\, e^{1-c}\right)}$$ and since $f(x)$ is increasing for any $x > x_*$, then $\cdots$.

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