The number of solution pairs (x, y) of the si | vedclass

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(B) Given equations are:$1) \log _{1 / 3}(x+y)+\log _3(x-y)=2$$2) 2^{y^2}=512^{x+1}$From equation $(1)$:$-\log _3(x+y)+\log _3(x-y)=2$$\log _3\left(\f

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$1$

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(B) Given equations are:$1) \log _{1 / 3}(x+y)+\log _3(x-y)=2$$2) 2^{y^2}=512^{x+1}$From equation $(1)$:$-\log _3(x+y)+\log _3(x-y)=2$$\log _3\left(\frac{x-y}{x+y}\right)=2$$\frac{x-y}{x+y}=3^2=9$$x-y=9x+9y$$-8x=10y \Rightarrow y = -\frac{4}{5}x$From equation $(2)$:$2^{y^2}=(2^9)^{x+1} = 2^{9(x+1)}$$y^2=9(x+1)$Substitute $y = -\frac{4}{5}x$ into $y^2=9(x+1)$:$\left(-\frac{4}{5}x\right)^2=9(x+1)$$\frac{16}{25}x^2=9x+9$$16x^2=225x+225$$16x^2-225x-225=0$$(16x+15)(x-15)=0$So,$x=15$ or $x=-\frac{15}{16}$.For $x=15$,$y=-\frac{4}{5}(15)=-12$. Check domain: $x+y=15-12=3 > 0$ and $x-y=15-(-12)=27 > 0$. This is a valid solution.For $x=-\frac{15}{16}$,$y=-\frac{4}{5}(-\frac{15}{16})=\frac{3}{4}$. Check domain: $x+y=-\frac{15}{16}+\frac{12}{16}=-\frac{3}{16} Thus,there is only $1$ solution pair $(15, -12)$.

The number of solution pairs $(x, y)$ of the simultaneous equations $\log _{1 / 3}(x+y)+\log _3(x-y)=2$ and $2^{y^2}=512^{x+1}$ is

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