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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

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

Commented by Tawa11 last updated on 30/Jun/22

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

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

$$\mathbb{T}\mathbf{han}\mathbb{k}\mathbf{s}\mathbf{miss}!$$

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