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I have two equations as below:

m = c - n m = log(n) + 1 

What approach should I take to solve this.

I am sorry, I forgot to mention the variables here.

'c' is a constant whose value is known. The equation is to be solved for 'm' and 'n'.

P.S. This is not a homework

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    possible duplicate of [Inverse of $y=xe^x$](http://math.stackexchange.com/questions/10261/inverse-of-y-xex). Put $\log n = t$. We get $t + 1 = c - e^t$. Now put $t+1 -c = -x$. We get $-x = e^{-x+c-1}$ i.e $xe^{x} = -e^{c-1}$.2011-06-06
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    I think this question is different enough from the linked question. I've flagged "exact duplicate" questions incredibly more similar to each other this post and your linked post, that didn't get closed. However, that's not to say that the linked post contains some information to may assist in solving this problem.2011-06-06
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    @Aryabhatta: now your link is more relevant. Why don't you post your comment as an answer/hint, together with linking the post you posted in your comment?2011-06-06
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    @SidCool: could you edit your question to indicate what it is you are to solve for? n? both n and c?2011-06-06
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    @amwhy: http://meta.math.stackexchange.com/questions/1756/coping-with-abstract-duplicate-questions2011-06-06
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    Thanks for reminding me @amWhy. I have edited the question.2011-06-06
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    @Aryabhatta: thanks for the link. I'm just confused at the lack of uniformity in what gets closed as "duplicate"...I see the connection here, though the user might not...in other cases, (solve for a side of a right triangle when the other sides are given) for which there are countless "duplicates"...I tried to flag one yesterday: no action taken? I"m not at all opposed to the philosophy of referring users to similar previously posted answers; I'd just like to see more consistency/clarity in what counts as "duplicate".2011-06-06
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    @amWhy: You can always discuss those of meta if you feel an action needs to be taken (btw which question was it?). It will surely happen that some questions will slip through, that is unavoidable. Also, what is a duplicate is subjective (for instance, it seems we disagree regarding this question) and that might explain the lack of uniformity.2011-06-06
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    @Aryabhatta: the question I flagged as a possible duplicate is at http://math.stackexchange.com/q/43510/9003, which, I just saw, was closed (12 hours after I flagged it)...I guess I have to be patient... I'd rather error on the side of allowing a questionable duplicate, but some are so obviously similar that it gets annoying. I guess, e.g. in a case like this one here, there are going to be people who don't recognize a connection to a linked "duplicate", whereas experienced math folks recognize patterns between questions in all the forms they may take, see conceptual similarities, etc.2011-06-06
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    @Aryabhatta: if I have further issues regarding what counts as duplicate, I'll be sure to post on meta...I'll delete my comments here shortly.2011-06-06

2 Answers 2

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Since $c-n = \log(n)+1$, $c-1 = n+\log(n)$, and in turn $e^{c-1}= n e^n$. Thus $n={\rm W}(e^{c-1})$ where W is Lambert W function.

EDIT: Adding details in response to the OP's request.

Put $x=e^{c-1} (> 0)$. Then, there exists a unique (real) number $n=n(x)$ such that $x=ne^n$. For example, if $c=2$, so that $x=e$, then $n=1$ (since $e=1e^1$). As another example, if $c=3$, so that $x=e^2$, then $n \approx 1.5571455989976$; indeed, letting $\hat n=1.5571455989976$, $\hat n e^{\hat n}$ is very close to $e^2$. The unique solution $n=n(x)$ of the equation $x=ne^n$ ($x > 0$) is given by (defined) $n={\rm W}(x)$, where ${\rm W}$ is the Lambert W function (thus, for example, ${\rm W}(e)=1$ and ${\rm W}(e^2) \approx 1.5571455989976$). While the function W is rather complicated, it can be evaluated immediately using WolframAlpha. So, given the constant $c$, just ask WolframAlpha to compute ${\rm W}(e^{c-1})(=n)$ (note that the function W is implemented as ProductLog).

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    I am sorry but I am no mathematician, can you give me some explanation on this?2011-06-07
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    @SidCool: Yes, I'll add some details later on.2011-06-07
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    details added...2011-06-07
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    Thanks Shai. Appreciate your help.2011-06-07
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Eliminating $m$, you have $c - n = \log(n) + 1$. Unfortunately, there is no "elementary" expression for a solution of this equation. You need to use the Lambert W function: $n = \hbox{LambertW}(exp(c-1))$.