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Find all $x, y \in \mathbb{R}$ such that: $$16^{x^2 + y} + 16^{x + y^2} = 1$$ The first obvious approach was to take the log base $16$ of both sides: $$\log_{16}(16^{x^2 + y} + 16^{x + y^2}) = 0...
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Question. Find all pairs (x,y) of real numbers such that 16x2+y +16x+y2 =1. Solution. Verified by Toppr. We have, 16x2+y +16x+y2 = 1. x2 +y+x+y2 + 1 2=(x+ 1 2)2 +(y+ 1 2)2 ≥ 0. This shows that x2 +y+x+y2 ≥ −1 2. By AM-GM inequality, 1 =16x2+y +16x+y2 ≥ 2(16x2+y.16x+y2)1 2. 2(16x2+y+x+y2)1 2 ≤ 2(16)−1 4 = 1 . This equality holds every where.
9 Ιουλ 2019 · En los problemas 1 a 40, determine la solución general de la ecuación diferencial dada. Especifique un intervalo en el cual esté definida la solución general. Linear Differential Equation How...
