Many equations cannot be solved exactly without using special functions. For example, to solve requires the use of the function (or similar). This function is sometimes defined in terms of an integral from which their properties can be deduced. Thus is defined by and it is then clear that, for example,
There are many equations that can only be solved in terms of newly-defined functions. One such function that isn’t all that well known is the LambertW function where is defined as a solution (for ) of . This allows you to solve equations like which was asked about on the S.O.S. Mathematics CyberBoard
To solve let so that . Then
Thus and so which is our answer.
Using tables or software this gives 1.100.
But hang on, is that the only solution? No, because for small values of and grows much faster than so for large values of . Since both and are continuous on there is another value of for which . A quick fiddle with a calculator gives .
Research into the LambertW function to find out how this other solution can be given in terms of this function.