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Question Number 172599 by Mikenice last updated on 29/Jun/22

Answered by Rasheed.Sindhi last updated on 29/Jun/22

8y^x −y^(2x) =16_((i)) , 2^x =y^2 _((ii)) ; x=?,y=? (i)⇒8(y^2 )^(x/2) −(y^2 )^x =16 8(2^x )^(x/2) −(2^x )^x =16 8∙2^(x^2 /2) −2^x^2 =16 Let 2^(x^2 /2) =t⇒2^x^2 =t^2 8t−t^2 =16 t^2 −8t+16=0 (t−4)^2 =0 t=4 2^(x^2 /2) =4=2^2 (x^2 /2)=2 x=±2 y^2 =2^(±2) y=±2^(±1) =±2,±(1/2) (x,y)=(2,2),(2,−2),(−2,(1/2)),(−2,−(1/2))

$$\underset{\left({i}\right)} {\underbrace{\mathrm{8}{y}^{{x}} −{y}^{\mathrm{2}{x}} =\mathrm{16}}}\:,\:\underset{\left({ii}\right)} {\underbrace{\mathrm{2}^{{x}} ={y}^{\mathrm{2}} }}\:;\:\:\:\:\:{x}=?,{y}=? \\ $$$$\left({i}\right)\Rightarrow\mathrm{8}\left({y}^{\mathrm{2}} \right)^{{x}/\mathrm{2}} −\left({y}^{\mathrm{2}} \right)^{{x}} =\mathrm{16} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{8}\left(\mathrm{2}^{{x}} \right)^{{x}/\mathrm{2}} −\left(\mathrm{2}^{{x}} \right)^{{x}} =\mathrm{16} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{8}\centerdot\mathrm{2}^{{x}^{\mathrm{2}} /\mathrm{2}} −\mathrm{2}^{{x}^{\mathrm{2}} } =\mathrm{16} \\ $$$$\:\:\:\:\:\:\:\:{Let}\:\:\mathrm{2}^{{x}^{\mathrm{2}} /\mathrm{2}} ={t}\Rightarrow\mathrm{2}^{{x}^{\mathrm{2}} } ={t}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{8}{t}−{t}^{\mathrm{2}} =\mathrm{16} \\ $$$$\:\:\:\:\:\:{t}^{\mathrm{2}} −\mathrm{8}{t}+\mathrm{16}=\mathrm{0} \\ $$$$\:\:\:\:\:\left({t}−\mathrm{4}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:{t}=\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\mathrm{2}^{{x}^{\mathrm{2}} /\mathrm{2}} =\mathrm{4}=\mathrm{2}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:{x}=\pm\mathrm{2} \\ $$$$\:\:\:\:\:\:{y}^{\mathrm{2}} =\mathrm{2}^{\pm\mathrm{2}} \\ $$$$\:\:\:\:\:\:{y}=\pm\mathrm{2}^{\pm\mathrm{1}} =\pm\mathrm{2},\pm\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left({x},{y}\right)=\left(\mathrm{2},\mathrm{2}\right),\left(\mathrm{2},−\mathrm{2}\right),\left(−\mathrm{2},\frac{\mathrm{1}}{\mathrm{2}}\right),\left(−\mathrm{2},−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$

Commented by Tawa11 last updated on 30/Jun/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by Rasheed.Sindhi last updated on 30/Jun/22

Thanks miss!

$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{miss}}! \\ $$

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