Question

$left { {{(frac{1}{9})^{-y} = 3^{2x-5}} atop {log_{2}(3y+8x-3)=log_{2}lg10000+log_{32}x^5}} right.$

 

Сложная система уравнений с логарифмами :((

Answer 1

$begin{cases} (frac{1}{9})^{-y}=3^{2x-5}\ log_2(3y+8x-3)=log_2lg10000+log_{32}x^5 end{cases}Rightarrow\ Rightarrow begin{cases} (3^{-2})^{-y}=3^{2x-5}\ log_2(3y+8x-3)=log_24+log_{2^5}x^5 end{cases}Rightarrow\ Rightarrow begin{cases} 3^{2y}=3^{2x-5}\ log_2(3y+8x-3)=log_24+frac55log_{2}x end{cases}Rightarrow\ Rightarow begin{cases} 2y=2x-5\ log_2(3y+8x-3)=log_24x end{cases}Rightarrow begin{cases} y=x-frac52\ 3left(x-frac52right)+8x-3=4x end{cases}$

$3left(x-frac52right)+8x-3=4x\ 3x-frac{15}2+8x-3-4x=0\ 7x=frac{21}2\ x=frac32=1,5\ begin{cases} y=-1\ x=1,5 end{cases}$