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Question Number 119815 by bemath last updated on 27/Oct/20
$${Given}\:{f}\left({x}+{y}\right)=\mathrm{4}{f}\left({x}\right).{f}\left({y}\right)\:{for} \\ $$$${all}\:{real}\:{numbers}\:{x}\:{and}\:{y}. \\ $$$${If}\:{f}\left(\mathrm{3}\right)=\mathrm{32}\:{then}\:{f}\left(\mathrm{1}\right)=\_ \\ $$
Answered by Olaf last updated on 27/Oct/20
$${f}\left(\mathrm{1}+\mathrm{1}\right)\:=\:\mathrm{4}{f}\left(\mathrm{1}\right){f}\left(\mathrm{1}\right) \\ $$$${f}\left(\mathrm{2}\right)\:=\:\mathrm{4}{f}\left(\mathrm{1}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${f}\left(\mathrm{2}+\mathrm{1}\right)\:=\:\mathrm{4}{f}\left(\mathrm{2}\right){f}\left(\mathrm{1}\right) \\ $$$${f}\left(\mathrm{3}\right)\:=\:\mathrm{16}{f}\left(\mathrm{1}\right)^{\mathrm{3}} \:=\:\mathrm{32} \\ $$$${f}\left(\mathrm{1}\right)\:=\:\sqrt[{\mathrm{3}}]{\mathrm{2}} \\ $$$$ \\ $$
Commented by bemath last updated on 27/Oct/20
$${yes}.... \\ $$