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I've been struggling with this one today. Applying the logarithmic function to both sides of the equation doesn't seem to work. Any ideas?

angryavian
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Rodrigo Almeida
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  • Hint: prove that the LHS is strictly increasing on $\mathbb{R}^+$, then look for an "easy" solution. – dxiv Mar 03 '17 at 04:58
  • As above, hint: $x=1/2$ (square root) – Zain Patel Mar 03 '17 at 04:59
  • I was actually looking for a more analytical solution. The numerical one itself isn't hard to find by inspection, but analytically it seems to be tricky. – Rodrigo Almeida Mar 03 '17 at 05:06
  • @RodrigoAlmeida The equation has no closed form solution in general (though monotonicity ensures that a solution exists and is unique). The choice of the RHS in this case makes it pretty obvious that the problem was about spotting the "easy" solution by inspection. – dxiv Mar 03 '17 at 05:15

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Because of $\enspace\displaystyle 2^\frac{1}{2}=(2^2)^\frac{1}{2^2}\enspace$ you can write

$$x^x\ln x = -\sqrt{2}\ln 2 = -\frac{2}{\sqrt{2}}\ln 2 = \left(\frac{1}{2}\right)^{\frac{1}{2}}\ln \frac{1}{2^2} = \left(\frac{1}{2^2}\right)^{\frac{1}{2^2}}\ln \frac{1}{2^2}$$

so that you get $\enspace\displaystyle x=\frac{1}{2^2}=0.25\enspace$ and therefore $\enspace\displaystyle x^{-2}=16$ .

user90369
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