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Is it possible to solve a function with both exponential and logarithm such as
$a x^2 - b.\log(x) = c $
in closed form; where $a,b,c$ are constants and $a>0$ and $b>0$?

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Too easy for MO ; try Lambert w-function –  Feldmann Denis Aug 28 '12 at 9:10
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Good question, wrong website. We're about research. You'll get a better reception at math.stackexchange.com –  Gerry Myerson Aug 28 '12 at 13:08
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@Tom and Gerry, thanks for the positive response. Well it did come up in my research and I thought this would be a good place to ask. I got a closed form solution on Wolfram Alpha but no clue how it was reached. –  pratikag Aug 28 '12 at 14:34
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I will honestly admit that I laughed at "Tom and Gerry". –  Vidit Nanda Aug 28 '12 at 22:00
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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 $x=\left( -\frac{b}{2a}W\left( -\frac{2a}{b}e^{-2c/b}\right) \right) ^{1/2}$. –  Américo Tavares Aug 29 '12 at 0:12
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