Question Number 215519 by walterpieuler last updated on 09/Jan/25
$$ \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{Find}}\:\left(\frac{\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{x}}}\right)^{\boldsymbol{\mathrm{log}}_{\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{y}}}} \frac{\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{x}}}\:} \boldsymbol{\mathrm{below}}: \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:+\:\boldsymbol{\mathrm{xy}}\:−\:\mathrm{3}\boldsymbol{\mathrm{y}}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 10/Jan/25
![2x^2 + xy − 3y^2 = 0 2((x/y))+1−3((y/x))=0 let (x/y)=a 2a+1−(3/a)=0 2a^2 +a−3=0 (a−1)(2a+3)=0 a=1,−(3/2) (x/y)=1,−(3/2) ((y/x))^(log_(x/y) (y/x) ) =((y/x))^(−1) =(x/y)=1,−(3/2) [∵ log_(x/y) (y/x)=log_(x/y) ((x/y))^(−1) =−1]](https://www.tinkutara.com/question/Q215521.png)
$$\:\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:+\:\boldsymbol{\mathrm{xy}}\:−\:\mathrm{3}\boldsymbol{\mathrm{y}}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\:\mathrm{2}\left(\frac{{x}}{{y}}\right)+\mathrm{1}−\mathrm{3}\left(\frac{{y}}{{x}}\right)=\mathrm{0} \\ $$$${let}\:\frac{{x}}{{y}}={a} \\ $$$$\mathrm{2}{a}+\mathrm{1}−\frac{\mathrm{3}}{{a}}=\mathrm{0} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} +{a}−\mathrm{3}=\mathrm{0} \\ $$$$\left({a}−\mathrm{1}\right)\left(\mathrm{2}{a}+\mathrm{3}\right)=\mathrm{0} \\ $$$${a}=\mathrm{1},−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\frac{{x}}{{y}}=\mathrm{1},−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\: \\ $$$$\left(\frac{\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{x}}}\right)^{\boldsymbol{\mathrm{log}}_{\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{y}}}} \frac{\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{x}}}\:} =\left(\frac{{y}}{{x}}\right)^{−\mathrm{1}} =\frac{{x}}{{y}}=\mathrm{1},−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\left[\because\:\boldsymbol{\mathrm{log}}_{\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{y}}}} \frac{\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{x}}}=\boldsymbol{\mathrm{log}}_{\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{y}}}} \left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{−\mathrm{1}} =−\mathrm{1}\right] \\ $$