Question

Помогите, пожалуйста, решить эту систему логарифмических уравнений. СРОЧНО, ПОДРОБНО! ДАЮ 100 БАЛЛОВ
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Answer 1

$\displaystyle\bf\\\left \{ {{4^{-y} \log_{2} x=4} \atop {\log_{2} x-2^{-2y} =4}} \right. \\\\\\\left \{ {{2^{-2y} \cdot\log_{2} x=4} \atop {\log_{2} x-2^{-2y} =4}} \right.$

Сделаем замену :

$\displaystyle\bf\\2^{-2y} =m, \ m > 0\\\\\log_{2} x=n \ \ , \ x > 0\\\\\\\left \{ {{m\cdot n=4} \atop {m+n=4}} \right. \\\\\\\left \{ {{m=4-n} \atop {(4-n)\cdot n=4}} \right. \\\\\\\left \{ {{m=4-n} \atop {n^{2} -4n+4=0}} \right. \\\\\\\left \{ {{m=4-n} \atop {\left(n-2)^{2} =0 }} \righ$

$\displaystyle\bf\\\left \{ {{m=4-n} \atop {n-2=0}} \right. \\\\\\\left \{ {{m=4-2} \atop {n=2}} \right.\\\\\\\left \{ {{m=2} \atop {n=2}} \right. \\\\\\\log_{2} x=2\\\\x=2^{2} =4\\\\\\2^{-2y} =2\\\\-2y=1\\\\y=-0,5\\\\\\Otvet \ : \ \Big(4 \ ; \ -0,5\Big)$