If $$4^{(x+y)} = 256$$ and $$(256)^{(x-y)} = 4$$, then what is the value of x and y?
Given : $$4^{(x+y)} = 256$$
=>Â $$4^{(x+y)} = 4^4$$
=> $$(x+y)=4$$ ----------(i)
Similarly, $$(256)^{(x-y)} = 4$$
=> $$4^{4(x-y)}=4^1$$
=> $$(x-y)=\frac{1}{4}$$ --------(ii)
Adding equations (i) and (ii), we get :
=> $$2x=4+\frac{1}{4}=\frac{17}{4}$$
=> $$x=\frac{17}{8}$$
Substituting it in equation (i), => $$y=4-\frac{17}{8}=\frac{32-17}{8}$$
=> $$y=\frac{15}{8}$$
=> Ans - (A)
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