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I've been at it for a while but I can't get it. Can anyone help?

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1 Answer 1

up vote 5 down vote accepted

We can first rewrite this to $x^{1/x} = \frac{3}{5}$, and taking logs gives us $\frac{1}{x} \ln(x) = \ln(3/5)$. Substituting $y = 1/x$ we get $y \ln(1/y) = \ln(3/5)$ or, by multiplying both sides by $-1$, $y \ln(y) = \ln(5/3)$. Then use the Lambert W-function (see e.g. Example 4) to get $$y = \frac{\ln(5/3)}{W(\ln(5/3))} \quad \rightarrow \quad x = \frac{W(\ln(5/3))}{\ln(5/3)} \approx 0.6995.$$

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Shouldn't it be $y \ln (1/y)$? –  Gigili Jun 2 '12 at 16:15
    
@Gigili: $\ln(1/y) = -\ln(y)$, and I moved the $-$ to the right hand side. But I'll add a note for clarity. –  TMM Jun 2 '12 at 16:17
    
Ah right. Thanks. –  Gigili Jun 2 '12 at 16:19
    
We haven't learned the product logs yet and that's why this question stumped me. But you cleared it up so thanks! :) –  David Jun 2 '12 at 16:23

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