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Question Number 158593 by mathlove last updated on 06/Nov/21

	Answered by cortano last updated on 07/Nov/21
	${ ((log _y (x)=log _x (2))),((log _x (y).log _x (y)=log _x (16))) :} ⇒[ log _x (y)]^2 =4log _x (2) let log _x (y)=u⇒log _y (x)=(1/u) ⇒u^2 =(4/u) ⇒u^3 =4 ⇒u=(4)^(1/3) ⇒log _x (y)=(4)^(1/3) ⇒y=x^(4)^(1/3) ⇒log_x^(4)^(1/3) (x)=log_x (2) ⇒(1/( (4)^(1/3) )) log_x (x)=log_x (2) ⇒2^(4)^(1/3) =x$
	${\begin{cases} \log_{y} (x) = \log_{x} (2) \\ \log_{x} (y) . \log_{x} (y) = \log_{x} (16) \end{cases}$ $\Rightarrow {[\log_{x} (y)]}^{2} = 4 \log_{x} (2)$ $l e t \log_{x} (y) = u \Rightarrow \log_{y} (x) = \frac{1}{u}$ $\Rightarrow u^{2} = \frac{4}{u} \Rightarrow u^{3} = 4$ $\Rightarrow u = \sqrt[3]{4} \Rightarrow \log_{x} (y) = \sqrt[3]{4}$ $\Rightarrow y = x^{\sqrt[3]{4}}$ $\Rightarrow \log_{x^{\sqrt[3]{4}}} (x) = \log_{x} (2)$ $\Rightarrow \frac{1}{\sqrt[3]{4}} \log_{x} (x) = \log_{x} (2)$ $\Rightarrow 2^{\sqrt[3]{4}} = x$
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