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Re: Which of the following equations is true for all positive va [#permalink]
1
This problem is testing our logic of Exponents and Roots. Let's try to understand each and every option in detail:

A. \(\sqrt{x} + \sqrt{y} = \sqrt{x+y}\)
Now, \(\sqrt{x} + \sqrt{y}\) cannot be merged to get = \(\sqrt{x+y}\)
Let's take an example to understand this. Let x = 4 and y = 9
\(\sqrt{x} + \sqrt{y}\) = \(\sqrt{4} + \sqrt{9}\) = 2+3 = 5
\(\sqrt{x+y} = \sqrt{4+9}\)= \(\sqrt{13}\) which is around 3.6
So, clearly \(\sqrt{x} + \sqrt{y} ≠ \sqrt{x+y}\)

B. \(\sqrt{x^4 y^{16}} = x^2 y^4\)
Let's simplify LHS (Left Hand Side)
\(\sqrt{x^4 y^{16}}\) = \(\sqrt{x^4}\) * \(\sqrt{y^{16}}\)
= \(x^\frac{4}{2}\) * \(y^\frac{16}{2}\) = \(x^2\) * \(y^8\)
Which is ≠ RHS (Right Hand Side)

C. \(( x \sqrt{y} )( y \sqrt{x} ) = x^2 y^2\)
Let's take LHS
\(( x \sqrt{y} )( y \sqrt{x} )\) = \(x*y^\frac{1}{2}\) * \(y*x^\frac{1}{2}\)
= \(x^(1 + \frac{1}{2}) *y^(1 + \frac{1}{2}) \)
= \(x^(\frac{3}{2}) * y^(\frac{3}{2})\)
which is ≠ RHS

D. \(y \sqrt{x}+ y \sqrt{x} = \sqrt{4 xy^2}\)
Let's take LHS
\(y \sqrt{x}+ y \sqrt{x}\)
Now, both the terms are same so we can add them like normal variables and have
\(2*y \sqrt{x}\)
Let's simplify RHS
\(\sqrt{4 xy^2}\) = \(\sqrt{2^2 xy^2}\)
= \(2*y\sqrt{x}\)
which is same as RHS

We do not need to solve for E as we already got D as answer but am solving just to explain the logic

E. \(( x^y )( y^y) = (xy)^{2y}\)
LHS = \(( x^y )( y^y) \)
Now We are multiplying two exponents with same power and different bases. So, we can multiply the bases and keep the power same
=> \(( x^y )( y^y) \) = \( (x*y)^y \)
which is not equal to RHS

So, answer will be D
Hope it helps!
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Re: Which of the following equations is true for all positive [#permalink]
Hello from the GRE Prep Club BumpBot!

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