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Question Number 121899 by ZiYangLee last updated on 12/Nov/20

Solve (log 2x)^2 +(log 3x)^2 =(log 2)^2 +(log 3)^2

$$\mathrm{Solve}\:\left(\mathrm{log}\:\mathrm{2}{x}\right)^{\mathrm{2}} +\left(\mathrm{log}\:\mathrm{3}{x}\right)^{\mathrm{2}} =\left(\mathrm{log}\:\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{log}\:\mathrm{3}\right)^{\mathrm{2}} \\ $$

Commented by Dwaipayan Shikari last updated on 12/Nov/20

x=1 2log^2 x+2log2logx+2log3logx=0 log^2 x+logxlog6=0 logx=−log6⇒x=(1/6) or logx=0 ,x=1

$${x}=\mathrm{1} \\ $$$$\mathrm{2}{log}^{\mathrm{2}} {x}+\mathrm{2}{log}\mathrm{2}{logx}+\mathrm{2}{log}\mathrm{3}{logx}=\mathrm{0} \\ $$$${log}^{\mathrm{2}} {x}+{logxlog}\mathrm{6}=\mathrm{0} \\ $$$${logx}=−{log}\mathrm{6}\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{6}}\:\:\:\:{or}\:{logx}=\mathrm{0}\:,{x}=\mathrm{1} \\ $$

Commented by ZiYangLee last updated on 12/Nov/20

thanks sir

$$\mathrm{thanks}\:\mathrm{sir} \\ $$

Answered by MJS_new last updated on 12/Nov/20

(log nx)^2 =(log x)^2 +2log n log x +(ln n)^2 transforming the given equation we get log x log (6x) =0 ⇒ x=1∨x=(1/6)

$$\left(\mathrm{log}\:{nx}\right)^{\mathrm{2}} =\left(\mathrm{log}\:{x}\right)^{\mathrm{2}} +\mathrm{2log}\:{n}\:\mathrm{log}\:{x}\:+\left(\mathrm{ln}\:{n}\right)^{\mathrm{2}} \\ $$$$\mathrm{transforming}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{log}\:{x}\:\mathrm{log}\:\left(\mathrm{6}{x}\right)\:=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\mathrm{1}\vee{x}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$

Commented by ZiYangLee last updated on 12/Nov/20

Thanks sir..★

$$\mathrm{Thanks}\:\mathrm{sir}..\bigstar \\ $$

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