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

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Answer 1 · 2017-11-25T22:41:21

$x = - W_{n} (- e^{- 2})$ $x = -W_n(-e^(-2))" "$ 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 $n in ZZ$

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