0
$\begingroup$

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 $b>0$?

$\endgroup$
8
  • 3
    $\begingroup$ Too easy for MO ; try Lambert w-function $\endgroup$ Aug 28, 2012 at 9:10
  • 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... $\endgroup$ Aug 28, 2012 at 14:17
  • 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. $\endgroup$ Aug 28, 2012 at 14:21
  • 3
    $\begingroup$ I will honestly admit that I laughed at "Tom and Gerry". $\endgroup$ Aug 28, 2012 at 22:00
  • 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 $x=\left( -\frac{b}{2a}W\left( -\frac{2a}{b}e^{-2c/b}\right) \right) ^{1/2}$. $\endgroup$ Aug 29, 2012 at 0:12

1 Answer 1

2
$\begingroup$

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 $x$ $$x=\Bigg( -\frac{b}{a n}\,W\Big(-\frac{a n}{b}e^{-\frac{c n}{b}} \Big) \Bigg)^{\frac{1}{n}}$$

$\endgroup$
1
  • 2
    $\begingroup$ This explanation is also in the comments, by @AméricoTavares (without the substitution of $n$ for $2$). $\endgroup$
    – LSpice
    Jan 18 at 15:30

Not the answer you're looking for? Browse other questions tagged or ask your own question.