How to solve $\ln(x) = 3\left(1-\frac{1}{x}\right)$?

Question

I have been working for this problem for a while:

$$\ln(x) = 3\left(1-\frac{1}{x}\right)$$

and by graphing, plugging and chugging values and rigorously doing the math, I can clearly see that one of the values that satisfy this condition is $x = 1$.

However, when I put this into wolfram alpha and Desmos, I see two answers, one that is $x = 1$ and another one that is approximately $16.801$.

It is expressed by the Lambert W Function. I solved for $x = 1$ by using the Lambert W function. I cannot find any way to solve for the latter solution: $16.801$.

Is there anyone that could elaborate for me how to solve for it? Thank you.

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My method to get $x=1$: $$\begin{align}\ln(x) = 3\left(1-\frac{1}{x}\right)\:&\Longrightarrow\:x=e^3e^{-\frac{3}{x}}\\&\Longrightarrow\:e^{-3}=\frac{1}{x}e^{-\frac{3}{x}}\\&\Longrightarrow\:-3e^{-3}=-\frac{3}{x}e^{-\frac{3}{x}}\end{align}$$

Applying Lambert W Function, which does the following: $W(ae^a) = a $:

$$-3 = -3/x \Longleftrightarrow x=1$$

Answer 1

$-3e^{-3}=(-3/x)e^{-3/x}$ as a graph

The Lambert W function has two real valued branches when x is between -1/e and 0 exclusive, which means it is not a function in this particular case because it is not 1-1. However, if you apply the other branch of the function you will get the other value ~16.801.

Answer 2

You correctly found : $$-3e^{-3} = (-3/x) e^{-3/x}$$ Of course an obvious first solution is $x=1$ .

Using the LambertW function : $$x\,e^x=a \quad\implies\quad x=W(a)$$

$$W(a)e^{W(a)}=a\quad\text{or equivalently}\quad x=W(x\,e^x)$$ Thus $$\begin{cases} a=-3e^{-3}\\ W(a)=-\frac{3}{x} \end{cases}$$ The solution is :$\quad x=-\frac{3}{W(a)}$ $$x=-\frac{3}{W(-3e^{-3})}\simeq 16.801016$$ Note :

The LambertW function is a multivalued function. The two real branches are noted $W_0(X)$ and $W_{-1}(X)$.

The above result corresponds to the branch $W_0$ which is commonly simply noted $W(X)$.

The second branch gives : $$x=-\frac{3}{W_{-1}(-3e^{-3})}=1.$$