Solve for x x^(log


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Question Number 124880 by WilliamsErinfolami last updated on 06/Dec/20

$S o l v e f o r x$ $x^{{l o g}_{5} 3 x} = 12 x$
	Answered by mr W last updated on 06/Dec/20

	$(\log_{5} 3 x) (\log_{5} x) = \log_{5} (12 x)$ $(\log_{5} 3 + \log_{5} x) (\log_{5} x) = \log_{5} 12 + \log_{5} x$ $(\log_{5} 3 + t) t = \log_{5} 12 + t$ $t^{2} - (1 - \log_{5} 3) t - \log_{5} 12 = 0$ $t = \frac{1 - \log_{5} 3 \pm \sqrt{{(1 - \log_{5} 3)}^{2} + {4 \log}_{5} 12}}{2}$ $\log_{5} x = \frac{1 - \log_{5} 3 \pm \sqrt{{(1 - \log_{5} 3)}^{2} + {4 \log}_{5} 12}}{2}$ $\Rightarrow x = 5^{\frac{1 - \log_{5} 3 \pm \sqrt{{(1 - \log_{5} 3)}^{2} + {4 \log}_{5} 12}}{2}}$ $= 9.693765, 0.171932$
	Answered by aleks041103 last updated on 06/Dec/20

	${l o g}_{x} (12 x) = {l o g}_{5} (3 x)$ $1 + {l o g}_{x} (12) = {l o g}_{5} 3 + {l o g}_{5} x$ $1 + \frac{{l o g}_{5} 12}{{l o g}_{5} x} = {l o g}_{5} 3 + {l o g}_{5} x$ $t = {l o g}_{5} x \Rightarrow x = 5^{t}$ $t + {l o g}_{5} 12 = t^{2} + ({l o g}_{5} 3) t$ $\Rightarrow t^{2} - (1 - {l o g}_{5} 3) t - {l o g}_{5} 12 = 0$ $t_{1, 2} = \frac{1 - {l o g}_{5} 3 \pm \sqrt{{(1 - {l o g}_{5} 3)}^{2} + 4 {l o g}_{5} 12}}{2}$ $\Rightarrow x_{1, 2} = 5^{t_{1, 2}}$
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