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Question Number 166633 by mathlove last updated on 23/Feb/22

Answered by mahdipoor last updated on 23/Feb/22

xyz=log_(2a) a×log_(3a) 2a×log_(4a) 3a=  ((lna)/(ln2a))×((ln2a)/(ln3a))×((ln3a)/(ln4a))=((lna)/(ln4a))  yz=log_(3a) 2a×log_(4a) 3a=((ln2a)/(ln3a))×((ln3a)/(ln4a))=((ln2a)/(ln4a))  1+xyz=((ln4a+lna)/(ln4a))=((ln4a^2 )/(ln4a))=((2ln2a)/(ln4a))=2yz

$${xyz}={log}_{\mathrm{2}{a}} {a}×{log}_{\mathrm{3}{a}} \mathrm{2}{a}×{log}_{\mathrm{4}{a}} \mathrm{3}{a}= \\ $$$$\frac{{lna}}{{ln}\mathrm{2}{a}}×\frac{{ln}\mathrm{2}{a}}{{ln}\mathrm{3}{a}}×\frac{{ln}\mathrm{3}{a}}{{ln}\mathrm{4}{a}}=\frac{{lna}}{{ln}\mathrm{4}{a}} \\ $$$${yz}={log}_{\mathrm{3}{a}} \mathrm{2}{a}×{log}_{\mathrm{4}{a}} \mathrm{3}{a}=\frac{{ln}\mathrm{2}{a}}{{ln}\mathrm{3}{a}}×\frac{{ln}\mathrm{3}{a}}{{ln}\mathrm{4}{a}}=\frac{{ln}\mathrm{2}{a}}{{ln}\mathrm{4}{a}} \\ $$$$\mathrm{1}+{xyz}=\frac{{ln}\mathrm{4}{a}+{lna}}{{ln}\mathrm{4}{a}}=\frac{{ln}\mathrm{4}{a}^{\mathrm{2}} }{{ln}\mathrm{4}{a}}=\frac{\mathrm{2}{ln}\mathrm{2}{a}}{{ln}\mathrm{4}{a}}=\mathrm{2}{yz} \\ $$

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