Question Number 172349 by Mikenice last updated on 25/Jun/22
Answered by mr W last updated on 26/Jun/22
$$\mathrm{log}\:\mathrm{3}\:×\left(\mathrm{log}\:\mathrm{3}+\mathrm{log}\:{x}\right)=\mathrm{log}\:\mathrm{4}\:×\left(\mathrm{log}\:\mathrm{4}+\mathrm{log}\:{y}\right) \\ $$$$\mathrm{log}\:\mathrm{4}\:×\mathrm{log}\:{x}=\mathrm{log}\:\mathrm{3}×\mathrm{log}\:{y} \\ $$$${let}\:{X}=\mathrm{log}\:{x},\:{Y}=\mathrm{log}\:{y} \\ $$$${a}=\mathrm{log}\:\mathrm{3},\:{b}=\mathrm{log}\:\mathrm{4} \\ $$$$\Rightarrow{a}\left({a}+{X}\right)={b}\left({b}+{Y}\right) \\ $$$$\Rightarrow{bX}={aY} \\ $$$$\Rightarrow{X}=−{a}\:\Rightarrow\mathrm{log}\:{x}=−\mathrm{log}\:\mathrm{3}\:\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{Y}=−{b}\:\Rightarrow\mathrm{log}\:{y}=−\mathrm{log}\:\mathrm{4}\:\Rightarrow{y}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$