Lambert-Function Solution of $\ln(x) = x-2$

Question

Found an equation on AP Calculus:$$ \ln(x) = x-2.$$ Two solutions: 0.15859, 3.14619. Got the first solution thus:

$e^{\ln(x)} = e^{x-2}$

$x = e^x e^{-2}$

$x (e^{-x}) = e^{-2}$

$-x (e^{-x}) = - e^{-2}$

$W[-x (e^{-x})]= W[- e^{-2}]$

$-x = W[- e^{-2}]$ .

Then I used productlog [- e^-2] on WolframAlpha.com to get 0.15859. Does anyone know how to get the other root, 3.14619?

(The people who originally solved this equation used a graphing calculator. Would rather do it the long way. Thanks.

Answer 1

You can use the second branch of the Lambert's function.

You can also notice that $|(\ln x)'|<1$ around that solution, so that the fixed-point iterations

$$x_{n+1}=\ln(x_n)+2$$ will converge.

Of course, Newton's iterations are much faster,

$$x_{n+1}=x_n-\frac{\ln x_n-x_n+2}{\dfrac{1}{x_n}-1}.$$