How do you solve #ln(x)=x-2#?

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Answer 1 · 2017-11-25T22:41:21

#x = -W_n(-e^(-2))" "# for any #n in ZZ#

The Lambert W function (actually a family of functions) satisfies:

#W_n(z e^z) = z#

Given:

#ln(x) = x - 2#

Taking the exponent of both sides, we get:

#x = e^(x-2)#

So:

#x e^(-x) = e^(-2)#

So:

#(-x) e^(-x) = -e^(-2)#

So:

#-x = W_n(-e^(-2))" "# for any branch of the Lambert W function.

So:

#x = -W_n(-e^(-2))" "# for any #n in ZZ#