Question 6

If $$16^{x}=\frac{64}{256^{y}}$$ and $$9^{x}=\frac{9}{3^{y}}$$, then the values of x and y are respectively,


From the first equation: $$16^x\cdot256^y=64$$ ==> $$4^{2x}\cdot4^{4y}=4^3$$ 
Comparing powers: 2x + 4y = 3 .....(A)

From second equation: $$3^{2x+y}=3^2$$
Comparing powers: 2x + y = 2 ......(B)

Solving A and B, we get option C as answer

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