Question Number 144609 by mathdanisur last updated on 26/Jun/21
$${How}\:{many}\:{digits}\:{doest}\:{the}\:{number} \\ $$$$\mathrm{2021}^{\mathrm{2022}} \:\:{have}.? \\ $$
Answered by MJS_new last updated on 27/Jun/21
$$\mathrm{log}_{\mathrm{10}} \:\mathrm{2021}^{\mathrm{2022}} \:=\mathrm{2022log}_{\mathrm{10}} \:\mathrm{2021}\:= \\ $$$$=\mathrm{2022log}_{\mathrm{10}} \:\left(\mathrm{2}.\mathrm{021}×\mathrm{10}^{\mathrm{3}} \right)\:=\mathrm{2022}\left(\mathrm{3}+\mathrm{log}_{\mathrm{10}} \:\mathrm{2}.\mathrm{021}\right)= \\ $$$$=\mathrm{6066}+\mathrm{2022log}_{\mathrm{10}} \:\mathrm{2}.\mathrm{021}\:\approx\mathrm{6683}.\mathrm{85508593}\approx \\ $$$$\approx\mathrm{7}.\mathrm{16285117718}×\mathrm{10}^{\mathrm{6683}} \:\mathrm{which}\:\mathrm{has}\:\mathrm{6684}\:\mathrm{digits} \\ $$
Commented by mathdanisur last updated on 27/Jun/21
$${alot}\:{cool}\:{thanks}\:{Sir} \\ $$