Question Number 196251 by sonukgindia last updated on 20/Aug/23
Answered by MrGHK last updated on 20/Aug/23
$${x}+{y}=\frac{{ln}\left(\mathrm{3}\right)}{{ln}\left(\mathrm{2}\right)} \\ $$$${x}−{y}=\frac{\mathrm{2}{ln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{3}\right)} \\ $$$$\left({x}+{y}\right)\left({x}−{y}\right)=\frac{{ln}\left(\mathrm{3}\right)}{{ln}\left(\mathrm{2}\right)}×\frac{\mathrm{2}{ln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{3}\right)} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{2} \\ $$
Answered by AST last updated on 20/Aug/23
$${x}−{y}=\mathrm{2}{log}_{\mathrm{3}} \mathrm{2};{x}+{y}={log}_{\mathrm{2}} \mathrm{3}\Rightarrow\left({x}+{y}\right)\left({x}−{y}\right)={x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{2} \\ $$
Commented by York12 last updated on 21/Aug/23
$${sir}\:{please}\:{look}\:{at}\: \\ $$$$\mathrm{196258} \\ $$
Answered by Rasheed.Sindhi last updated on 22/Aug/23
$$\mathrm{2}^{{x}+{y}} =\mathrm{3},\:\mathrm{3}^{{x}−{y}} =\mathrm{4};\:\:\:{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =? \\ $$$$\mathrm{3}^{{x}−{y}} =\mathrm{4}\Rightarrow\left(\mathrm{2}^{{x}+{y}} \right)^{{x}−{y}} =\mathrm{2}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{2} \\ $$