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How can I solve this for $x$?

$xe^x=-2/a$ with $(a \in \mathbb{R_0^+})$

$a$ can be any strict positive real number.

I need this because I'm searching for the root of a function to sketch a graph.

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    You can't do it with the usual elementary functions. But see the Wikipedia page on the [Lambert $W$-function](http://en.wikipedia.org/wiki/Lambert_W_function), especially the section on [applications](https://secure.wikimedia.org/wikipedia/en/wiki/Lambert_W_function#Applications) (J.M. or someone else will certainly jump in with some better links shortly). Concerning your notation: $(a \in \mathbb{R_0^+})$ means $a \gt 0$?2011-08-24
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    @Theo: I mean a has to be a strictly positive real number, so yes. However, I had to find this during my exam today, which is only in real numbers.. the link you provided uses complex numbers.2011-08-24
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    Strange... I would read that as $a \geq 0$. Anyway: Concerning your task in the exam: did you have to *solve* for $x$ or did you have to show that a solution $x$ *exists*? The latter can be achieved using the [intermediate value theorem](http://en.wikipedia.org/wiki/Intermediate_value_theorem).2011-08-24
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    Strictly positive, so not zero. The question was sketch a graph for $$f(x) = \dfrac{something}{axe^x + 2}$$. Maybe I misunderstood the question, and was I searching for the wrong things, but I believe you have to find the domain to sketch the graph. So I searched for roots of the denominator.2011-08-24
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    I see. I'm sure somebody will help you with that. Unfortunately I don't have time right now. You may want to change your question in view of your last comment, because answerers will undoubtedly work something out using Lambert $W$ the way you ask it. If you want to know how to sketch the graph, in your last comment, you should say so. (notice the *sketch* which doesn't ask you to determine the root in the denominator *exactly*).2011-08-24
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    Unfortunately, I forgot the whole function.. so asking for a graph of something-something seems a bit strange.2011-08-24
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    Well, why not put *something* $= 1$? That should tell you how to do it.2011-08-24
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    @TheoBuehler let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/1174/discussion-between-mats-and-theo-buehler)2011-08-24
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    Mats, the reals are a subset of the complex numbers so there's no issue. You can see what $W$ looks like on $\mathbb{R}$ on the graphic at the top of the linked Wikipedia page. The equation $$xe^x = b$$ **only** has real solution(s) if $b\ge -e^{-1}$. Also, nobody takes the chat warning seriously. :)2011-08-24

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I suspect anon's comment is what your exam paper was looking for: for $a$ sufficiently small, $ax e^x + 2 > 0$ always, so for a sketch of the graph you have something smooth with certain asymptotic behaviour.

For $a = 2e$, there is exactly one root, and you should show that you get blow-up of the same sign when you approach the singular point from the left and when you approach the singular point from the right. (Informally: $\lim_{x\to -1^+} (axe^x + 2)^{-1} = \lim_{x\to -1^-} (axe^x + 2)^{-1} = +\infty$)

For $a > 2e$, there are two roots, and at those two singularities, the left limit and the right limit have opposite signs, so you need to show that on the graph. In other words, I suspect the exam paper was asking for a qualitative depiction of the graph of the function, and not for a strictly quantitative depiction.