Editing Lambert W function (section)

=== Solving equations ===
The Lambert {{mvar|W}} function is used to solve equations in which the unknown quantity occurs both in the base and in the exponent, or both inside and outside of a logarithm. The strategy is to convert such an equation into one of the form {{math|1=''ze''<sup>''z''</sup> = ''w''}} and then to solve for {{mvar|z}} using the {{mvar|W}} function.

For example, the equation
: <math>3^x=2x+2</math>
(where {{mvar|x}} is an unknown real number) can be solved by rewriting it as
: <math>\begin{align} &(x+1)\ 3^{-x}=\frac{1}{2} & (\mbox{multiply by } 3^{-x}/2) \\
\Leftrightarrow\ &(-x-1)\ 3^{-x-1} = -\frac{1}{6} & (\mbox{multiply by } {-}1/3) \\
\Leftrightarrow\ &(\ln 3) (-x-1)\ e^{(\ln 3)(-x-1)} = -\frac{\ln 3}{6} & (\mbox{multiply by } \ln 3)
\end{align}</math>

This last equation has the desired form and the solutions for real ''x'' are:
: <math>(\ln 3) (-x-1) = W_0\left(\frac{-\ln 3}{6}\right) \ \ \ \textrm{ or }\ \ \ (\ln 3) (-x-1) = W_{-1}\left(\frac{-\ln 3}{6}\right) </math>
and thus:
: <math>x= -1-\frac{W_0\left(-\frac{\ln 3}{6}\right)}{\ln 3} = -0.79011\ldots \ \ \textrm{ or }\ \ x= -1-\frac{W_{-1}\left(-\frac{\ln 3}{6}\right)}{\ln 3} = 1.44456\ldots</math>

Generally, the solution to
: <math>x = a+b\,e^{cx}</math>
is:
: <math>x=a-\frac{1}{c}W(-bc\,e^{ac})</math>
where ''a'', ''b'', and ''c'' are complex constants, with ''b'' and ''c'' not equal to zero, and the ''W'' function is of any integer order.