Is it possible to solve a function with both exponential and logarithm such as
$$
a x^2 - b\cdot\log(x) = c
$$
in closed form; where $a,b,c$ are constants and $a>0$ and $b>0$?
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3$\begingroup$ Too easy for MO ; try Lambert w-function $\endgroup$– Feldmann DenisCommented Aug 28, 2012 at 9:10
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3$\begingroup$ I think this is one of those interesting cases where it's only clear to those who already know the answer that the question is not suitable for MO. Many professional mathematicians won't have heard of the Lambert w-function. If you haven't heard of it then of course you'll have no way of guessing that that's what you need to look up in order to solve equations like this. Nicely worded as Gerry's comment is, I actually disagree with the implication that this isn't a research question. I think it's a question that could easily come up in anyone's research... $\endgroup$– Tom LeinsterCommented Aug 28, 2012 at 14:17
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3$\begingroup$ ...and I continue to believe (perhaps swimming against the tide) that MO should welcome questions by researchers wanting to broaden out into fields other than their own. $\endgroup$– Tom LeinsterCommented Aug 28, 2012 at 14:21
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3$\begingroup$ I will honestly admit that I laughed at "Tom and Gerry". $\endgroup$– Vidit NandaCommented Aug 28, 2012 at 22:00
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5$\begingroup$ Short explanation. If you substitute $x=\sqrt{t}$ in the equation $ax^{2}-b\log x=c$ you can rewrite it as $-\frac{2a}{b}e^{-2c/b}=\left( -\frac{2a}{b}t\right) e^{-2at/b}$. Since by definition of the Lambert $W$ function $Y=Xe^{X}$ iff $X=W(Y)$ this means that $W\left( -\frac{2a}{b}e^{-2c/b}\right) =-\frac{2a}{b}t=-\frac{2a}{b}x^{2}$. And solving for $x$ you get $x=\left( -\frac{b}{2a}W\left( -\frac{2a}{b}e^{-2c/b}\right) \right) ^{1/2}$. $\endgroup$– Américo TavaresCommented Aug 29, 2012 at 0:12
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1 Answer
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As @AméricoTavares did in comments, making the problem more general $$a x^n - b\log(x) = c\qquad \implies \qquad a x^n - \frac bn \log(x^n) = c$$ $$x^n=t\quad \implies \quad a t - \frac bn \log(t) = c\quad \implies \quad t=-\frac{b}{a n}\,W\Big(-\frac{a n}{b}e^{-\frac{c n}{b}} \Big)$$
The solution is reeal if $$\frac{a n}{b}e^{-\frac{c n}{b}} \leq \frac 1e$$ Back to $x$ $$x=\Bigg( -\frac{b}{a n}\,W\Big(-\frac{a n}{b}e^{-\frac{c n}{b}} \Big) \Bigg)^{\frac{1}{n}}$$
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