Compare it: (log_4 20)^2 and log


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Question Number 158814 by HongKing last updated on 09/Nov/21

$Compare it :$ ${(\log_{4} 20)}^{2} and \log_{4} 320$
	Answered by Rasheed.Sindhi last updated on 09/Nov/21

	${(\log_{4} 20)}^{2} \overset{?}{=} \log_{4} 320$ ${(\log_{4} (2^{2} . 5))}^{2} \overset{?}{=} \log_{4} (2^{6} . 5)$ ${({2 \log}_{4} 2 + \log_{4} 5)}^{2} \overset{?}{=} {6 \log}_{4} 2 + \log_{4} 5$ ${(2 (\frac{1}{2}) + \log_{4} 5)}^{2} \overset{?}{=} 6 (\frac{1}{2}) + \log_{4} 5$ ${(1 + \log_{4} 5)}^{2} \overset{?}{=} 3 + \log_{4} 5$ $1 + {(\log_{4} 5)}^{2} + {2 \log}_{4} 5 \overset{?}{=} 3 + \log_{4} 5$ ${(\log_{4} 5)}^{2} + \log_{4} 5 \overset{?}{=} 2$ $O b v i o u s l y \log_{4} 5 > \log_{4} 4 = 1$ ${(\log_{4} 5)}^{2} + \log_{4} 5 > {(\log_{4} 4)}^{2} + \log_{4} 4$ ${(\log_{4} 5)}^{2} + \log_{4} 5 > 1 + 1 = 2$ $∴ {(\log_{4} 20)}^{2} > \log_{4} 320$
	Commented by HongKing last updated on 09/Nov/21

	$thank you very much my dear Ser cool$
	Answered by Raxreedoroid last updated on 09/Nov/21

	$\log_{4} 20 = 1 + \log_{4} 5$ ${(\log_{4} 20)}^{2} = \log_{4} 20 + (\log_{4} 20) (\log_{4} 5)$ $\log_{4} 320 = \log_{4} 20 + \log_{4} 16$ $∴ {(\log_{4} 20)}^{2} > \log_{4} 320$
	Commented by HongKing last updated on 09/Nov/21

	$thank you very much dear Ser cool$
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