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Question Number 114561 by Akeyz last updated on 19/Sep/20

	Answered by bemath last updated on 19/Sep/20
	$log _((log _π x)) (log _x π) = −1=log _((log _π x)) (log _π x)^(−1) log _x π = log _π x ((ln π)/(ln x)) = ((ln x)/(ln π)) → { ((ln x=ln π⇒x=π)),((ln x=ln ((1/π))⇒x=(1/π))) :}$
	$\log_{(\log_{π} x)} (\log_{x} π) = - 1 = \log_{(\log_{π} x)} {(\log_{π} x)}^{- 1}$ $\log_{x} π = \log_{π} x$ $\frac{\ln π}{\ln x} = \frac{\ln x}{\ln π} \to {\begin{cases} \ln x = \ln π \Rightarrow x = π \\ \ln x = \ln (\frac{1}{π}) \Rightarrow x = \frac{1}{π} \end{cases}$
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