Given \(x, y > 0\) & asked is \(81^{3} = 27^{(y-x)}\)?
Reduce the equation in question to a simpler form, by reducing it to a common base.
\(81^{3} =( 3^{4})^{3} = 3^{12}\)
&
\(27^{(y-x)} = 3^{3(y-x)}\)
So we have, \(3^{12} = 3^{3(y-x)}\)
hence, \(12 = 3(y-x)\)
simplified to \(4 = y-x\), hence our rephrased question is to find out whether \((y-x) = 4\) ?
Stmt 1: \(/ y-x / = 4\)
that means \((y-x) = 4\) or \((y-x) = -4\)
hence
Stmt 1 alone is Insufficient. Therefore Choice A & D are not possible.
Stmt 2: \(2^{y} = 8^{x}\)
reduced to \(2^{y} = 2^{3x}\)
hence \(y = 3x\)
so, \((y-x) = 3x - x = 2x\)
given \(x>0\) , hence x can take any positive value.
hence
Stmt 2 alone is Insufficient. Therefore Choice B is not possible.
Combining Stmt 1 & 2, we get,
\((y-x) = 4\) or \(-4\)
&
\((y-x) =2x\)
Therefore,
\(2x = 4\) or \(-4\)
\(x = 2\) or \(-2\)
however \(x>0\), hence \(x=2\), therefore \((y-x)=4\)
So combining
Stmt 1 & 2 is sufficient, hence
C is the answer choice.Thanks,
Gym
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