I think Example 1 in https://en.wikipedia.org/wiki/Lambert_W_function#Example_1 might be applicable to this! For instance, for a x + b = ln x, set y = ln x - b to get (a e^b) e^y = y. Then -a e^b = -y e^(-y), and so y = -W(-a e^b). Substituting back then gives x = e^(b-W(-a e^b)), assuming I’ve done my math right! Mathematica informs me that since a x + b = ln x, this can also be written as x = -W(-a e^b)/a. --Neil Bickford On Mon, Aug 12, 2019 at 2:11 AM Andres Valloud < avalloud@smalltalk.comcastbiz.net> wrote:
What's a neat way to solve equations such as the below exactly?
ax + b = ln x
In particular, what to do when b = 0?
Andres. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun