Question

Найдите меньший корень уравнения 4^2x+1 - 7*12^x +3^2x+1 =0

Answer 1

$4^{2x+1} -7*12^{x}+3^{2x+1}=0\\\\4*4^{2x}-7*12^{x}+3*3^{2x}=0\\\\4*16^{x} -7*12^{x}+3*9^{x}=0|:9^{x}>0\\\\4*(\frac{16}{9})^{x}-7*(\frac{12}{9})^{x}+3=0\\\\4*(\frac{4}{3})^{2x}-7*(\frac{4}{3})^{x}+3=0\\\\(\frac{4}{3})^{x}=m,m>0\\\\4m^{2}-7m+3=0\\\\D=(-7)^{2}-4*4*3=49-48=1\\\\m_{1}=\frac{7+1}{8}=1\\\\m_{2}=\frac{7-1}{8}=\frac{6}{8}=\frac{3}{4}$

$1)(\frac{4}{3})^{x}=1\\\\(\frac{4}{3})^{x}=(\frac{4}{3})^{o}\\\\x_{1}=0\\\\2)(\frac{4}{3})^{x}=\frac{3}{4}\\\\(\frac{4}{3})^{x}=(\frac{4}{3})^{-1}\\\\x_{2}=-1\\\\Otvet:\boxed{-1}$