logx_(4x) +logx_(x/2) =2 solve for x=?


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Question Number 161786 by mathlove last updated on 22/Dec/21

$l o g \underset{4 x}{x} + l o g \underset{\frac{x}{2}}{x} = 2$ $s o l v e f o r x = ?$
	Answered by mr W last updated on 22/Dec/21

	$\frac{\ln x}{\ln (4 x)} + \frac{\ln x}{\ln (\frac{x}{2})} = 2$ $\frac{\ln x}{2 \ln 2 + \ln x} + \frac{\ln x}{\ln x - \ln 2} = 2$ $l e t t = \ln x, a = \ln 2$ $\frac{t}{2 a + t} + \frac{t}{t - a} = 2$ $t = 4 a$ $\ln x = 4 \ln 2$ $\Rightarrow x = 2^{4} = 16$
	Answered by cortano last updated on 22/Dec/21

	$(Q) S o l v e f o r x : \log_{4 x} (x) + \log_{\frac{x}{2}} (x) = 2$ $(S) \Rightarrow \frac{\log_{2} (x)}{2 + \log_{2} (x)} + \frac{\log_{2} (x)}{\log_{2} (x) - 1} = 2$ $[z = \log_{2} (x); x > 0; x \neq 1]$ $\Rightarrow \frac{z}{z + 2} + \frac{z}{z - 1} = 2$ $\Rightarrow z (z - 1 + z + 2) = 2 (z + 2) (z - 1)$ $\Rightarrow 2 z^{2} + z = 2 z^{2} + 2 z - 4$ $\Rightarrow z = 4 = \log_{2} (x)$ $\Rightarrow x = 2^{4} = 16$
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