Предмет: Математика, автор: ismika

Решить систему x^(log 2 (y))+y^(log 2 (x))=16; log 2 (x)-log 2 (y)=2

Ответы

Автор ответа: Minsk00
2
Решить систему
 \left \{ {{x^{log_2(y)}+y^{log_2(x)}=16}} \atop {log_2 (x)-log_2 (y)=2}} \right.
Решение
Область допустимых значений системы уравнений 
x>0, y>0.
Для наглядности решения заменим переменные
x=2^t                     y=2^z
log₂(x)=t                                     log₂(y)=z
Подставим в исходную систему уравнений
 \left \{ {{(2^t)^z+(2^z)^t=16}} \atop {t-z=2}} \right.
 \left \{ {{2^{tz}+2^{tz}=16}} \atop {t-z=2}} \right.\ \textless \ =\ \textgreater \ \left \{ {{2*2^{tz}=16}} \atop {t-z=2}} \right.\ \textless \ =\ \textgreater \ \left \{ {{2^{tz}=8}} \atop {t-z=2}} \right.\ \textless \ =\ \textgreater \  \left \{ {{2^{tz}=2^3}} \atop {t-z=2}} \right.
\left \{ {{tz=3}} \atop {t-z=2}} \right.
Решаем систему уравнений по методу подстановки
Из второго уравнения выразим переменную t и подставим в первое уравнение
                                     t =2+z
                             (2+z)z=3
                           z
²+2z-3=0
D=2²-4*(-3)=4+12=16
                        z₁ = (-2 - 4)/2 = -3          t₁ = 2 - 3 = -1
                        z₂ = (-2 + 4)/2 = 1          t₂ = 2 + 1 = 3
Делаем обратную замену и находим переменные х и у
При 
z₁=-3          t₁=-1
                   x₁ = 2^t = 2^(-1) = 1/2          y₁ = 2^z = 2^(-3) = 1/8       
При z₂=1          t₂=3
                   x₂ = 2^t = 2^(3) = 8               y₂ = 2^z = 2^(1) = 2
 Получили две пары ответов 
        х=8  у=2
Проверка
 \left \{ {{8^{log_2(2)}+2^{log_2(8)}=8^1+2^3=8+8=16}} \atop {log_2 (8)-log_2 (2)=log_2 (2^3)-1=3-1=2}} \right.
и
       х=1/2=0,5   у=1/8=0,125
Проверка
 \left \{ {{ (\frac{1}{2}) ^{log_2( \frac{1}{8} )}+ (\frac{1}{8}) ^{log_2( \frac{1}{2} )}= \frac{1}{2} ^{-3}+ \frac{1}{8} ^{-1}=8+8=16}} \atop {log_2 ( \frac{1}{2} )-log_2 ( \frac{1}{8} )=log_2 (2^{-1})-log_2 (2)^{-3}=-1+3=2}} \right.

 Ответ: (8;2); (0,5;0,125) 
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