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Question Number 11005 by ajfour last updated on 06/Mar/17

$p r o v e t h a t \frac{9}{4} < {(\log_{2} 3)}^{2} < \frac{25}{9} .$
	Answered by mrW1 last updated on 06/Mar/17

	${2 \log}_{2} 3 = \log_{2} 3^{2} = \log_{2} 9 > \log_{2} 8 = \log_{2} 2^{3} = 3$ $\Rightarrow \log_{2} 3 > \frac{3}{2}$ $\Rightarrow {(\log_{2} 3)}^{2} > {(\frac{3}{2})}^{2} = \frac{9}{4}$ ${3 \log}_{2} 3 = \log_{2} 3^{3} = \log_{2} 27 < \log_{2} 32 = \log_{2} 2^{5} = 5$ $\Rightarrow \log_{2} 3 < \frac{5}{3}$ $\Rightarrow {(\log_{2} 3)}^{2} < {(\frac{5}{3})}^{2} = \frac{25}{9}$ $\Rightarrow \frac{9}{4} < {(\log_{2} 3)}^{2} < \frac{25}{9}$
	Commented by ajfour last updated on 07/Mar/17

	$T h a n k s, s o e a s y f o r y o u!$
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