<s>[INST] Why does $\sqrt{x-x^{\frac{1}{x}^{\frac{1}{x}}}}=\log_{\sqrt{x-x^{\frac{1}{x}^{\frac{1}{x}}}}}(x)?$ (Error) 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 $\sqrt{x-x^{\frac{1}{x}^{\frac{1}{x}}}}=\log_{\left(\sqrt{x-x^{\frac{1}{x}^{\frac{1}{x}}}}\right)^{-1}}(x)$!!
 [/INST] 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 $(a^b)^c = a^{bc}$.
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