If log _3 x+log _3 y=2+log _3 2 and log _3(x+ | vedclass

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(C) Given equations are: $1$) $\log _3 x + \log _3 y = 2 + \log _3 2$ $2$) $\log _3(x+y) = 2$ From equation $(1)$, using the property $\log_a b + \log

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$x=3, y=6$

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(C) Given equations are: $1$) $\log _3 x + \log _3 y = 2 + \log _3 2$ $2$) $\log _3(x+y) = 2$ From equation $(1)$, using the property $\log_a b + \log_a c = \log_a(bc)$: $\log _3(xy) = \log _3(3^2) + \log _3 2$ $\log _3(xy) = \log _3 9 + \log _3 2$ $\log _3(xy) = \log _3(9 	imes 2) = \log _3 18$ So, $xy = 18$. From equation $(2)$, using the definition of logarithm: $x + y = 3^2 = 9$. We have a system of equations: $x + y = 9$ $xy = 18$ These are the roots of the quadratic equation $t^2 - (x+y)t + xy = 0$: $t^2 - 9t + 18 = 0$ $(t-3)(t-6) = 0$ So, $t = 3$ or $t = 6$. Thus, the solutions are $(x=3, y=6)$ or $(x=6, y=3)$. Comparing with the given options, the correct choice is $x=3, y=6$.

If $\log _3 x+\log _3 y=2+\log _3 2$ and $\log _3(x+y)=2$, then

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