Question Number 19171 by gourav~ last updated on 06/Aug/17
$${log}_{\sqrt{\mathrm{2}}} \:\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\:\:\:\:}}}} \\ $$
Commented by gourav~ last updated on 06/Aug/17
$${find}\:{value} \\ $$
Answered by Tinkutara last updated on 06/Aug/17
$$\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}}}}}\:=\:\mathrm{2}^{\frac{\mathrm{2}^{\mathrm{4}} \:−\:\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }} \:=\:\mathrm{2}^{\frac{\mathrm{15}}{\mathrm{16}}} \:=\:\left(\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{15}}{\mathrm{8}}} \\ $$$$\mathrm{log}_{\sqrt{\mathrm{2}}} \left(\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{15}}{\mathrm{8}}} \:=\:\frac{\mathrm{15}}{\mathrm{8}} \\ $$
Commented by Tinkutara last updated on 06/Aug/17
$$\mathrm{See}\:\mathrm{Q}.\:\mathrm{18551}. \\ $$
Commented by chernoaguero@gmail.com last updated on 06/Aug/17
$${sir}\:{how}\:{did}\:{u}\:\frac{\mathrm{2}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }\:\:{plz}\:{be}\:{simplistic} \\ $$