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$p = \lambda e^{ -\lambda }$

If $p$ and $e$ are known. How can I calculate $\lambda$?

I tried on log on both side but it did not help. Any suggestions?

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You say that $e$ is known -- does that mean you're using $e$ as a variable name rather than to denote Euler's number? That would be rather confusing.

In either case, what you need is the Lambert W function. Using the defining relation

$z=W(z)\mathrm e^{W(z)}$

(where $\mathrm e$ as usual denotes Euler's number) and

$-p=-\lambda\mathrm e^{-\lambda}\;,$

you get

$\lambda =-W(-p)\;.$

Note that this is a) multivalued and b) only defined for $p\le1/e$.

If you really did intend to use $e$ as a variable name, we can replace it by $a$ to keep things clear and then write

$ \begin{eqnarray} p &=& \lambda a^{-\lambda} \;, \\ p &=& \lambda \mathrm e^{-\lambda\log a} \;, \\ -p\log a &=& -\lambda\log a \mathrm e^{-\lambda\log a} \;, \\ \lambda &=& -\frac{W(-p\log a)}{\log a} \;. \end{eqnarray}$

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    @Patrick: Thanks -- already corrected; you were $f$aster :-)2011-12-07