Question : If $4^{(x+y)} = 256$ and $(256)^{(x–y)} = 4$, what are the values of $x$ and $y?$
Option 1: $\frac{17}{8}$ and $\frac{15}{8}$
Option 2: $\frac{17}{4}$ and $\frac{15}{4}$
Option 3: $\frac{9}{17}$ and $\frac{15}{17}$
Option 4: $\frac{8}{17}$ and $\frac{8}{15}$
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Correct Answer: $\frac{17}{8}$ and $\frac{15}{8}$
Solution :
Given: $4^{(x+y)}=256$
⇒ $4^{(x+y)} = 4^4$
⇒ $x+y = 4$
⇒ $x = 4-y$-----(1)
And $(256)^{(x-y)}=4$
⇒ $(4^4)^{(x-y)}=4^1$
⇒ $4x-4y =1$-----(2)
Putting the value of $x$ from equation (1), we get,
$4(4-y)-4y = 1$
⇒ $16-4y-4y =1$
$\therefore y =\frac{15}{8}$
Now, substituting the value of $y$ in equation (1), we get,
$x = 4-\frac{15}{8}$
$\therefore x = \frac{17}{8}$
Hence, the correct answer is $\frac{17}{8}$ and $\frac{15}{8}$.
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