x-2-xy-4x-y-2-xy-4y-log-16-x-1-y-1-x-2-y-2-

Question Number 149040 by mathdanisur last updated on 02/Aug/21

{\begin{cases} x^{2} + x y = 4 x \\ y^{2} + x y = 4 y \end{cases} \Rightarrow {l o g}_{16}^{(x_{1} + y_{1} + x_{2} + y_{2})} = ?

Answered by bramlexs22 last updated on 02/Aug/21

(1) + (2) \Leftrightarrow x^{2} + 2 xy + y^{2} = 4 (x + y)

\Rightarrow {(x + y)}^{2} = 4 (x + y); (x + y) (x + y - 4) = 0

case (1) \Rightarrow x = - y

\Rightarrow x^{2} - x^{2} = 4 x \Rightarrow {\begin{cases} x_{1} = 0 \\ y_{1} = 0 \end{cases}

case (2) x = 4 - y

\Rightarrow x (x + y) = 4 x; 4 x = 4 x \to x_{2} = λ, y_{2} = 4 - λ

then \log_{16} (x_{1} + y_{1} + x_{2} + y_{2}) = \log_{16} (0 + λ + 4 - λ)

= \log_{16} (4) = \frac{1}{2}

Commented by mathdanisur last updated on 02/Aug/21

T h a n k Y o u S e r, C o o l

T h a n k Y o u S e r, C o o l