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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

$$={\log }_{\mathrm{n}}2+{2\log }_{\mathrm{n}}2+{3\log }_{\mathrm{n}}2+...+{2015\log }_{\mathrm{n}}2$$$$={\log }_{\mathrm{n}}2(1+2+3+...+2015)$$$$={\log }_{\mathrm{n}}2.\frac{2016.2015}{2}=2015.\frac{9!}{6!3!}$$$$\Rightarrow {\log }_{\mathrm{n}}2=\frac{1}{12}\therefore {\log }_2\mathrm{n}=12\Rightarrow \mathrm{n}=2^{12}$$$$$$

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

$$\mathrm{Thanks}.$$

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