Question Number 225837 by Rojarani last updated on 14/Nov/25
$${Show}\:{that},\:{log}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}….\alpha}}}}\:=\mathrm{1} \\ $$
Answered by fantastic last updated on 14/Nov/25
$${let}\:\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}..\infty}}}}}={x} \\ $$$${x}^{\mathrm{2}} =\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}..\infty}}}}} \\ $$$$\mathrm{7}{x}=\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}\sqrt{\mathrm{7}..\infty}}}}} \\ $$$${x}^{\mathrm{2}} =\mathrm{7}{x}\Rightarrow{x}=\mathrm{7}\:{orx}=\mathrm{0} \\ $$$${logx}={log}\mathrm{7} \\ $$$${log}\mathrm{0}={not}\:{possible} \\ $$
Answered by mr W last updated on 14/Nov/25
$$=\mathrm{log}\:\mathrm{7}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}+…+\frac{\mathrm{1}}{\mathrm{2}^{{n}} }+…} \\ $$$$=\mathrm{log}\:\mathrm{7}^{\mathrm{1}} =\mathrm{log}\:\mathrm{7}\neq\mathrm{1} \\ $$
Answered by AgniMath last updated on 15/Nov/25
$${there}\:{should}\:{be}\:\mathrm{7}\:{at}\:{the}\:{base} \\ $$