I am just wondering if there is an easy way to solve
$$\frac{\log(x)}{x}=c, \text{where } x > 1 \text{ and } c < e^{-1}$$
Asked
Active
Viewed 71 times
1 Answers
3
No, there isn't. The solution is $$x=-\frac{W(-c)}{c},$$ where $W$ is the Lambert W function, and in the solution $W$ means both $W_0$ and $W_{-1}$.
To give more details.
$$ \begin{align} \frac{\log\left(-\frac{W(-c)}{c}\right)}{-\frac{W(-c)}{c}} & = c \\ \log(-W(-c))-\log c & = -W(-c) \\ \log(-W(-c))+W(-c) & = \log c \\ W(-c)e^{W(-c)} & = -c, \end{align} $$ and this satisfies the definition of Lambert W function.
user153012
- 12,240
-
Then the remaining question is: Is this easy? I.e., is the answer to the OP “yes” or “no”? I suppose that depends on perspective … – Harald Hanche-Olsen Sep 04 '14 at 19:27
-
Two solutions possible, I think. Check sign changes at $x=1, \infty$. – Macavity Sep 04 '14 at 19:32
-
@Macavity Right! Do you know how to find the other solution? – Elrond Gimli Sep 04 '14 at 19:34
-
1The second real soln is $-W_{-1}(-c)/c$, on the other side of $e$. – Macavity Sep 04 '14 at 19:38
-
@Macavity Thanks! Does it matter if I use $W_{-1}(c)/c$ or $W_{-1}(-c)/-c$ for the second answer? – Elrond Gimli Sep 04 '14 at 19:44
-
Yes, it does. Since $W_{-1}(c)/c \neq -W_{-1}(-c)/c$. – user153012 Sep 04 '14 at 19:55