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solved using auxiliary variable (so they ask) I can not build the auxiliary variable for this problem, if they ask

log in base 10

$$10^{\log ( \log x )}-10^{\log (16/\log x)}=6$$

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$$ 10^{\log ( \log x )}-10^{\log (16/\log x)}=6\Rightarrow 10^{\log ( \log x )}-10^{\log 16}10^{-\log(\log x)}=6\ . $$ Then set $10^{\log(\log x)}=t$, so your equation becomes $$ t-10^{\log 16}\frac{1}{t}=6\Rightarrow t^2-6t-10^{\log 16}=0\Rightarrow t=-2\text{ or }t=8\ . $$ We need to exclude $t=-2$, so the equation to be solved is $$ 10^{\log(\log x)}=8\Rightarrow 10^{\log(\log x)}=10^{\log 8}\Rightarrow \log(\log x)=\log 8\Rightarrow \log x=8\Rightarrow \boxed{x=10^8}\ . $$

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  • $\begingroup$ Thanks, sorry I had not seen, thanks for your time $\endgroup$ – zeros Jun 27 '16 at 23:21

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