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