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Re: Which of the following equations is true for all positive va
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07 Jun 2020, 02:40
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
Re: Which of the following equations is true for all positive
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15 May 2024, 09:52
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Re: Which of the following equations is true for all positive [#permalink]