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I might be being very silly here, but I can't for the life of me see why $$\sqrt{x-x^{\frac{1}{x}^{\frac{1}{x}}}}=\log_{\sqrt{x-x^{\frac{1}{x}^{\frac{1}{x}}}}}(x)$$for $x\in \mathbb{Z}, x>1$?

$\left(\text{ie, }\log\text{ to the base }\sqrt{x-x^{\frac{1}{x}^{\frac{1}{x}}}}.\right)$

Update

As pointed out by almagest, Did & GEdgar, this is incorrect. I had a "$1/\dots$" hiding in the formula, so of course $\sqrt{x-x^{\frac{1}{x}^{\frac{1}{x}}}}=\log_{\left(\sqrt{x-x^{\frac{1}{x}^{\frac{1}{x}}}}\right)^{-1}}(x)$!!

Community
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martin
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1 Answers1

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Write $$ y = \sqrt{x-x^{\frac{1}{x}^{\frac{1}{x}}}} $$ Now $y=\log_y x$ means $y^y=x$. As almagest noted, try $x=2$, so that $$ y = \sqrt{2-2^{\frac{1}{2}^{\frac{1}{2}}}} = \sqrt{2-2^{1/\sqrt{2}}} \approx 0.606 $$ and $$ y^y \approx 0.738, $$ and this result is not $x=2$.

That's with the (standard) interpretation $a^{b^c} = a^{(b^c)}$. It also fails with the other interpretation $(a^b)^c = a^{bc}$.

GEdgar
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