Question

Помогите решить уравнения

Answer 1

23)

$\frac{2^{x}+10 }{4}=\frac{9}{2^{x-2}}\\\\\frac{2^{x}+10 }{4}=\frac{9}{2^{x}*\frac{1}{4}}\\\\\frac{2^{x}+10 }{4}=\frac{36}{2^{x}}|*4*2^{x}\\\\(2^{x})^{2}+10*2^{x}=144\\\\2^{x}=m,m>0$

$m^{2}+10m-144=0\\\\D=10^{2}-4*(-144)=100+576=676=26^{2}\\\\m_{1}=\frac{-10+26}{2}=8\\\\m_{2}=\frac{-10-26}{2}=-18<0\\\\2^{x}=8\\\\2^{x}=2^{3}\\\\x=3\\\\Otvet:3$

27)

$log_{3}(3^{x^{2}-13x+28} +\frac{2}{9})=log_{5}0,2\\\\3^{x^{2} -13x+28}+\frac{2}{9}>0\\\\log_{3}(3^{x^{2}-13x+28 }+\frac{2}{9})=-1\\\\3^{x^{2}-13x+28 }+\frac{2}{9}=\frac{1}{3}$

$3^{x^{2}-13x+28 }=\frac{1}{9}\\\\3^{x^{2}-13x+28 }=3^{-2}\\\\x^{2}-13x+28=-2\\\\x^{2}-13x+30=0\\\\x_{1}=3\\\\x_{2}=10\\\\Otvet:3;10$