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Wondering if there’s a closed form for $\exp(\frac{1}{\log(x)})=\frac{x}{e}$?

Numerically solving yields $x \approx .539$

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  • $\begingroup$ WolframAlpha says, that also $x \approx 5.04317$ is a solution. $\endgroup$ – Viktor Glombik Dec 12 '19 at 21:16
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$\exp \left (\dfrac{1}{\ln x} \right ) = \dfrac{x}{e}; \tag 1$

$\dfrac{1}{\ln x} = \ln \left ( \dfrac{x}{e} \right ) = \ln x - \ln e = \ln x - 1; \tag 2$

$1 = (\ln x)^2 - \ln x; \tag 3$

$(\ln x)^2 - \ln x - 1 = 0; \tag 4$

quadratic formula:

$\ln x = \dfrac{1 \pm \sqrt 5}{2}; \tag 5$

$x = \exp \left ( \dfrac{1 \pm \sqrt 5}{2} \right ). \tag 6$

Some Additional Observations:

From (5),

$\dfrac{1}{\ln x} = \dfrac{2}{1 \pm \sqrt 5} = \dfrac{2(1 \mp \sqrt 5)}{(1 \mp \sqrt 5)(1 \pm \sqrt 5}$ $= \dfrac{2(1 \mp \sqrt 5)}{1^2 - (\sqrt 5)^2} = \dfrac{2(1 \mp \sqrt 5)}{1 - 5} = \dfrac{2(1 \mp \sqrt 5)}{-4} = - \dfrac{1 \mp \sqrt 5}{2}; \tag 7$

and also from (5)

$\ln x - 1 = \dfrac{1 \pm \sqrt 5}{2} - 1 = \dfrac{-1 \pm \sqrt 5}{2} = - \dfrac{1 \mp \sqrt 5}{2} = \dfrac{1}{\ln x}; \tag 8$

from (6),

$\dfrac{x}{e} = xe^{-1} = \exp \left ( \dfrac{1 \pm \sqrt 5}{2} \right )e^{-1} = \exp \left ( \dfrac{1 \pm \sqrt 5}{2} - 1 \right )$ $= \exp \left ( -\dfrac{1 \mp \sqrt 5}{2} \right ) = \exp \left ( \dfrac{1}{\ln x} \right ); \tag 9$

we also note that (4) yields

$\ln x( \ln x - 1) = 1 \Longrightarrow \dfrac{1}{\ln x} = \ln x - 1, \tag{10}$

in agreement with (8).

The reader is advised to take some care in keeping careful track of the $\pm$ and $\mp$ signs occurring in the above, since I have taken some liberty in their use. But the intended meaning shouldn't be too hard to discern.

The reader may recall that the quadratic equation

$\phi^2 - \phi - 1 = 0, \tag{11}$

whose roots are of course

$\phi = \dfrac{1 \pm \sqrt 5}{2}, \tag{12}$

and obey

$\dfrac{1}{\phi} = \phi - 1, \tag{13}$

in fact quantifies the cassical golden section, which is the ratio of the sides of a rectangle such that if a square whose side is the shorter is removed the remaining rectangle is in the same proportion as the original. So apparently what we're looking at here is the exponential/logarithmic version of that. The further pursuit of this correspondence fascinates, but will be deferred to a later time.

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    $\begingroup$ HI, approved every thing: the question and the answers. :-) My best regards. $\endgroup$ – Sebastiano Dec 12 '19 at 8:40
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    $\begingroup$ @Sebastiano: thank you my friend. ;) Happy Holidays and, of course, Cheers! $\endgroup$ – Robert Lewis Dec 12 '19 at 19:01
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On rewriting the equation,

$$\frac{1}{\ln x} = \ln\frac{x}{e} = \ln x - \ln e = \ln x -1$$ $$(\ln x)^2 - \ln x - 1 = 0$$ $$\ln x = \frac{1\pm\sqrt{1+4}}{2} = \frac{1\pm\sqrt5}{2}$$

$$x = \exp\left( \frac{1\pm\sqrt5}{2}\right)$$

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    $\begingroup$ $+1$ good spot. OP's numerical answer is the $(-)$ root $\endgroup$ – Ninad Munshi Dec 12 '19 at 8:15
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    $\begingroup$ Thank You! Yeah. $\endgroup$ – Ak19 Dec 12 '19 at 8:16

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