The number of integral solutions of the equation 4log_(x/2) ((√x))+2log_(4x) (x^2 )= 3log


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Question Number 22220 by Tinkutara last updated on 13/Oct/17

$The number of integral solutions of the$ $equation {4 \log}_{x / 2} (\sqrt{x}) + {2 \log}_{4 x} (x^{2}) =$ ${3 \log}_{2 x} (x^{3}) is$
	Answered by ajfour last updated on 13/Oct/17

	$4 \log_{x / 2} [\sqrt{2} (\sqrt{x / 2})] + 2 \log_{4 x} [\frac{{(4 x)}^{2}}{16}]$ $= 3 \log_{2 x} [\frac{{(2 x)}^{3}}{8}]$ $\Rightarrow 4 \log_{x / 2} \sqrt{2} + 4 (\frac{1}{2}) + 2 \times 2 - 2 \log_{4 x} 16$ $= 3 \times 3 - 3 \log_{2 x} 8$ $\frac{4 \log_{2} \sqrt{2}}{\log_{2} x - 1} - \frac{2 \log_{2} 16}{\log_{2} x + 2} = 3 - \frac{3 \log_{2} 8}{\log_{2} x + 1}$ $l e t \log_{2} x = t, t h e n$ $\frac{2}{t - 1} - \frac{8}{t + 2} = 3 - \frac{9}{t + 1}$ $i f t \neq 1, - 2, - 1 t h e n$ $\frac{2}{t - 1} + \frac{9}{t + 1} = 3 + \frac{8}{t + 2} \Rightarrow$ $\frac{2 t + 2 + 9 t - 9}{t^{2} - 1} = \frac{3 t + 6 + 8}{t + 2}$ $(11 t - 7) (t + 2) = (t^{2} - 1) (3 t + 14)$ $11 t^{2} + 15 t - 14 = 3 t^{3} + 14 t^{2} - 3 t - 14$ $3 t^{3} + 3 t^{2} - 18 t = 0$ $\Rightarrow t (t^{2} + t^{2} - 6) = 0$ $o r t (t + 3) (t - 2) = 0$ $t = \log_{2} x = 0, 2, - 3$ $\Rightarrow x = 1, 4, \frac{1}{8}$ $h e n c e t w o i n t e g r a l s o l u t i o n s .$
	Commented by Tinkutara last updated on 14/Oct/17

	$Thank you very much Sir!$
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