Question Number 207250 by hardmath last updated on 10/May/24
$$\mathrm{If}\:\:\:\mathrm{2}^{\boldsymbol{\mathrm{a}}} \:=\:\mathrm{5}\:\:,\:\:\mathrm{3}^{\boldsymbol{\mathrm{b}}} \:=\:\mathrm{9}\:\:\mathrm{and}\:\:\mathrm{25}^{\boldsymbol{\mathrm{c}}} \:=\:\mathrm{8} \\ $$$$\mathrm{Find}:\:\:\mathrm{a}\centerdot\mathrm{b}\centerdot\mathrm{c}\:=\:? \\ $$
Answered by A5T last updated on 10/May/24
$${a}={log}_{\mathrm{2}} \mathrm{5};{b}=\mathrm{2};{c}={log}_{\mathrm{25}} \mathrm{8}=\frac{\mathrm{1}}{\mathrm{2}}{log}_{\mathrm{5}} \mathrm{8}={log}_{\mathrm{5}} \left(\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$$\Rightarrow{abc}=\mathrm{2}{log}_{\mathrm{2}} \mathrm{5}×{log}_{\mathrm{5}} \mathrm{2}\sqrt{\mathrm{2}}={log}_{\mathrm{2}} \mathrm{8}=\mathrm{3} \\ $$
Commented by hardmath last updated on 10/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Answered by Rasheed.Sindhi last updated on 11/May/24
$$\mathrm{3}^{\boldsymbol{\mathrm{b}}} \:=\:\mathrm{9}\Rightarrow\mathrm{3}^{\mathrm{b}} =\mathrm{3}^{\mathrm{2}} \Rightarrow\mathrm{b}=\mathrm{2} \\ $$$$\mathrm{2}^{\mathrm{a}} =\mathrm{5}\Rightarrow\mathrm{2}=\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{a}}} \\ $$$$\mathrm{25}^{\mathrm{c}} =\mathrm{8}\Rightarrow\mathrm{5}^{\mathrm{2c}} =\mathrm{2}^{\mathrm{3}} =\left(\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{a}}} \right)^{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{5}^{\mathrm{2c}} =\mathrm{5}^{\frac{\mathrm{3}}{\mathrm{a}}} \Rightarrow\mathrm{2c}=\frac{\mathrm{3}}{\mathrm{a}}\Rightarrow\mathrm{ac}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left(\mathrm{ac}\right)\left(\mathrm{b}\right)=\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{2}\right)=\mathrm{3}\Rightarrow\mathrm{abc}=\mathrm{3} \\ $$