Question

Решить уравнение
4х-3×2 в степени х =4

Answer 1

$4^{x}-3\cdot2^{x}=4\\\\(2^{x})^{2}-3\cdot2^{x}-4=0\\\\2^{x} =m \ , \ m>0\\\\m^{2}-3m-4=0\\\\D=(-3)^{2} -4\cdot(-4)=9+16=25=5^{2}\\\\m_{1} =\dfrac{3+5}{2}=4\\\\m_{2}=\frac{3-5}{2}=-1<0-neyd\\\\2^{x} =4\\\\2^{x} =2^{2}\\\\Otvet:\boxed{x=2}$

Answer 2

$\displaystyle\bf \boxed{ \star t=2^x \quad;\quad t^2=4^x \star} \quad ODZ: t>0 \\\\4^x-3\cdot2^x=4 \\t^2-3t=4\\t^2-3t-4=0 \\\\\left \{ {{t_1+t_2=3} \atop{t_1\cdot t_2=-4}} \right. => \boxed{t_1=4} \quad ; \quad t_2\neq -1 \quad; ODZ : t>0 \\\\2^x=4 => x=2 \\\\Otvet: \boxed{x=2}$