Question Number 191833 by Mingma last updated on 01/May/23
Answered by AST last updated on 01/May/23
$${a}={log}_{\mathrm{2}} \mathrm{900};{b}={log}_{\mathrm{3}} \mathrm{900};{c}={log}_{\mathrm{5}} \mathrm{900} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}={log}_{\mathrm{900}} \mathrm{2}+{log}_{\mathrm{900}} \mathrm{3}+{log}_{\mathrm{900}} \mathrm{5} \\ $$$$={log}_{\mathrm{900}} \left(\mathrm{2}×\mathrm{3}×\mathrm{5}\right)={log}_{\mathrm{900}} \left(\mathrm{30}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by BaliramKumar last updated on 01/May/23
$$\mathrm{2}^{{a}} \:=\:\mathrm{900} \\ $$$$\mathrm{3}^{{b}\:} =\:\mathrm{900} \\ $$$$\mathrm{5}^{{c}} \:=\:\mathrm{900} \\ $$$$\mathrm{2}^{{a}} ×\mathrm{3}^{{b}} ×\mathrm{5}^{{c}} \:=\:\mathrm{900}×\mathrm{900}×\mathrm{900}\:=\:\left(\mathrm{30}^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$$\mathrm{2}^{{a}} ×\mathrm{3}^{{b}} ×\mathrm{5}^{{c}} \:=\:\mathrm{30}^{\mathrm{6}} \:=\:\left(\mathrm{2}×\mathrm{3}×\mathrm{5}\right)^{\mathrm{6}} \\ $$$$\mathrm{2}^{{a}} ×\mathrm{3}^{{b}} ×\mathrm{5}^{{c}} \:=\:\mathrm{2}^{\mathrm{6}} ×\mathrm{3}^{\mathrm{6}} ×\mathrm{5}^{\mathrm{6}} \\ $$$${a}\:=\:{b}\:=\:{c}\:=\:\mathrm{6} \\ $$$$\frac{\mathrm{6}×\mathrm{6}+\mathrm{6}×\mathrm{6}+\mathrm{6}×\mathrm{6}}{\mathrm{6}×\mathrm{6}×\mathrm{6}}\:=\:\frac{\mathrm{3}×\cancel{\mathrm{6}}×\cancel{\mathrm{6}}}{\mathrm{6}×\cancel{\mathrm{6}}×\cancel{\mathrm{6}}}\:=\:\frac{\mathrm{3}}{\mathrm{6}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{answer}\:\mathrm{is}\:\mathrm{right}\:\mathrm{by}\:\mathrm{mistake} \\ $$
Commented by Tinku Tara last updated on 01/May/23
$$\mathrm{2}^{\mathrm{6}} \neq\mathrm{900} \\ $$$$\mathrm{2}^{{a}} \mathrm{3}^{{b}} \mathrm{5}^{{c}} =\mathrm{30}^{\mathrm{6}} \nRightarrow{a}={b}={c}=\mathrm{6} \\ $$
Commented by BaliramKumar last updated on 01/May/23
$$\mathrm{sorry}\:\mathrm{i}\:\mathrm{ignore}\:\mathrm{first}\:\mathrm{three}\:\mathrm{equ}. \\ $$
Answered by HeferH last updated on 01/May/23
$$\mathrm{2}^{\mathrm{a}} =\:\mathrm{900} \\ $$$$\mathrm{3}^{\mathrm{b}} =\:\mathrm{900} \\ $$$$\mathrm{5}^{\mathrm{c}} \:=\:\mathrm{900} \\ $$$$\:\mathrm{2}\centerdot\mathrm{3}\centerdot\mathrm{5}\:=\:\mathrm{900}^{\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}+\frac{\mathrm{1}}{\mathrm{c}}} \\ $$$$\:\mathrm{30}\:=\:\mathrm{30}^{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}+\frac{\mathrm{1}}{\mathrm{c}}\right)} \\ $$$$\:\therefore\:\frac{\mathrm{ab}+\mathrm{bc}+\mathrm{ac}}{\mathrm{abc}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$