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Compare-it-log-4-20-2-and-log-4-320-

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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
$$\mathrm{Compare}\:\mathrm{it}: \\ $$$$\left(\mathrm{log}_{\mathrm{4}} \mathrm{20}\right)^{\mathrm{2}} \:\:\:\mathrm{and}\:\:\:\mathrm{log}_{\mathrm{4}} \mathrm{320} \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 09/Nov/21
(log_4 20)^2 =^(?) log_4 320 (log_4 (2^2 .5))^2 =^(?) log_4 (2^6 .5) (2log_4 2+log_4 5 )^2 =^(?) 6log_4 2+log_4 5 (2((1/2))+log_4 5 )^2 =^(?) 6((1/2))+log_4 5 (1+log_4 5 )^2 =^(?) 3+log_4 5 1+(log_4 5)^2 +2log_4 5 =^(?) 3+log_4 5 (log_4 5)^2 +log_4 5 =^(?) 2 Obviously 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
$$\left(\mathrm{log}_{\mathrm{4}} \mathrm{20}\right)^{\mathrm{2}} \:\:\:\overset{?} {=}\:\:\:\mathrm{log}_{\mathrm{4}} \mathrm{320} \\ $$$$\left(\mathrm{log}_{\mathrm{4}} \left(\mathrm{2}^{\mathrm{2}} .\mathrm{5}\right)\right)^{\mathrm{2}} \:\:\:\overset{?} {=}\:\:\:\mathrm{log}_{\mathrm{4}} \left(\mathrm{2}^{\mathrm{6}} .\mathrm{5}\right) \\ $$$$\left(\mathrm{2log}_{\mathrm{4}} \mathrm{2}+\mathrm{log}_{\mathrm{4}} \mathrm{5}\:\:\:\right)^{\mathrm{2}} \overset{?} {=}\mathrm{6log}_{\mathrm{4}} \mathrm{2}+\mathrm{log}_{\mathrm{4}} \mathrm{5}\: \\ $$$$\left(\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{log}_{\mathrm{4}} \mathrm{5}\:\:\:\right)^{\mathrm{2}} \overset{?} {=}\mathrm{6}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{log}_{\mathrm{4}} \mathrm{5}\: \\ $$$$\left(\mathrm{1}+\mathrm{log}_{\mathrm{4}} \mathrm{5}\:\:\:\right)^{\mathrm{2}} \overset{?} {=}\mathrm{3}+\mathrm{log}_{\mathrm{4}} \mathrm{5}\: \\ $$$$\mathrm{1}+\left(\mathrm{log}_{\mathrm{4}} \mathrm{5}\right)^{\mathrm{2}} +\mathrm{2log}_{\mathrm{4}} \mathrm{5}\:\:\:\:\overset{?} {=}\mathrm{3}+\mathrm{log}_{\mathrm{4}} \mathrm{5}\: \\ $$$$\left(\mathrm{log}_{\mathrm{4}} \mathrm{5}\right)^{\mathrm{2}} +\mathrm{log}_{\mathrm{4}} \mathrm{5}\:\:\:\:\overset{?} {=}\mathrm{2} \\ $$$${Obviously}\:\mathrm{log}_{\mathrm{4}} \mathrm{5}>\mathrm{log}_{\mathrm{4}} \mathrm{4}=\mathrm{1} \\ $$$$\left(\mathrm{log}_{\mathrm{4}} \mathrm{5}\right)^{\mathrm{2}} +\mathrm{log}_{\mathrm{4}} \mathrm{5}>\left(\mathrm{log}_{\mathrm{4}} \mathrm{4}\right)^{\mathrm{2}} +\mathrm{log}_{\mathrm{4}} \mathrm{4} \\ $$$$\left(\mathrm{log}_{\mathrm{4}} \mathrm{5}\right)^{\mathrm{2}} +\mathrm{log}_{\mathrm{4}} \mathrm{5}>\mathrm{1}+\mathrm{1}=\mathrm{2} \\ $$$$\therefore\:\left(\mathrm{log}_{\mathrm{4}} \mathrm{20}\right)^{\mathrm{2}} \:\:\:>\:\:\:\mathrm{log}_{\mathrm{4}} \mathrm{320} \\ $$
Commented by HongKing last updated on 09/Nov/21
thank you very much my dear Ser cool
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser}\:\mathrm{cool} \\ $$
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
$$\mathrm{log}_{\mathrm{4}} \:\mathrm{20}\:=\mathrm{1}+\mathrm{log}_{\mathrm{4}} \:\mathrm{5} \\ $$$$\left(\mathrm{log}_{\mathrm{4}} \:\mathrm{20}\right)^{\mathrm{2}} =\mathrm{log}_{\mathrm{4}} \:\mathrm{20}+\left(\mathrm{log}_{\mathrm{4}} \:\mathrm{20}\right)\left(\mathrm{log}_{\mathrm{4}} \mathrm{5}\right) \\ $$$$\mathrm{log}_{\mathrm{4}} \:\mathrm{320}=\mathrm{log}_{\mathrm{4}} \:\mathrm{20}\:+\mathrm{log}_{\mathrm{4}} \:\mathrm{16} \\ $$$$\therefore\:\left(\mathrm{log}_{\mathrm{4}} \:\mathrm{20}\right)^{\mathrm{2}} >\mathrm{log}_{\mathrm{4}} \:\mathrm{320} \\ $$
Commented by HongKing last updated on 09/Nov/21
thank you very much dear Ser cool
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{Ser}\:\mathrm{cool} \\ $$

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