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Question Number 112538 by Aina Samuel Temidayo last updated on 08/Sep/20

Answered by floor(10²Eta[1]) last updated on 08/Sep/20

=log_n 2+2log_n 2+3log_n 2+...+2015log_n 2  =log_n 2(1+2+3+...+2015)  =log_n 2.((2016.2015)/2)=2015.((9!)/(6!3!))  ⇒log_n 2=(1/(12))∴log_2 n=12⇒n=2^(12)

$$=\mathrm{log}_{\mathrm{n}} \mathrm{2}+\mathrm{2log}_{\mathrm{n}} \mathrm{2}+\mathrm{3log}_{\mathrm{n}} \mathrm{2}+...+\mathrm{2015log}_{\mathrm{n}} \mathrm{2} \\ $$$$=\mathrm{log}_{\mathrm{n}} \mathrm{2}\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+...+\mathrm{2015}\right) \\ $$$$=\mathrm{log}_{\mathrm{n}} \mathrm{2}.\frac{\mathrm{2016}.\mathrm{2015}}{\mathrm{2}}=\mathrm{2015}.\frac{\mathrm{9}!}{\mathrm{6}!\mathrm{3}!} \\ $$$$\Rightarrow\mathrm{log}_{\mathrm{n}} \mathrm{2}=\frac{\mathrm{1}}{\mathrm{12}}\therefore\mathrm{log}_{\mathrm{2}} \mathrm{n}=\mathrm{12}\Rightarrow\mathrm{n}=\mathrm{2}^{\mathrm{12}} \\ $$$$ \\ $$

Commented by Aina Samuel Temidayo last updated on 08/Sep/20

Thanks.

$$\mathrm{Thanks}. \\ $$

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