Question

Решите уравнения 2*2^(2x)-17*2^x+8=0; 5*2^(2x)-7*10^x+2*5^(2x)=0 тема показательные уравнения

Answer 1

$2\cdot 2^{2x}-17\cdot 2^x+8=0~~~~~\Big|~~2^x=y0\\\\2y^2-17y+8=0\\D=17^2-4\cdot 2\cdot 8=289-64=225=15^2\\\\y_1=\dfrac{17+15}{2\cdot 2}=8=2^3~~~\Rightarrow~~2^x=2^3;~~\boldsymbol{x_1=3}\\\\y_2=\dfrac{17-15}{2\cdot 2}=\dfrac 12=2^{-1}~~~\Rightarrow~~2^x=2^{-1};~~\boldsymbol{x_2=-1}\\\\\boxed{\boldsymbol{x_1=3;~~~x_2=-1}}$

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$5\cdot 2^{2x}-7\cdot 10^x+2\cdot 5^{2x}=0\\5\cdot 2^{2x}-7\cdot 2^x\cdot 5^x+2\cdot 5^{2x}=0~~~~~\big|~~:(5^{2x})\neq 0\\\\5\cdot \dfrac{2^{2x}}{5^{2x}}-7\cdot \dfrac{2^x\cdot 5^x}{5^{2x}}+2\cdot \dfrac{5^{2x}}{5^{2x}}=0\\\\5\cdot \bigg(\dfrac 25\bigg)^{2x}-7\cdot \bigg(\dfrac 25\bigg)^x+2=0~~~~~\bigg|~~\bigg(\dfrac 25\bigg)^x=z0$

$5z^2-7z+2=0\\D=7^2-4\cdot 5\cdot 2=49-40=9=3^2\\\\z_1=\dfrac{7+3}{2\cdot 5}=1=\bigg(\dfrac 25\bigg)^0~~~\Rightarrow~~~\bigg(\dfrac 25\bigg)^x=\bigg(\dfrac 25\bigg)^0;~~\boldsymbol{x_1=0}\\\\z_2=\dfrac{7-3}{2\cdot 5}=\dfrac 25~~~\Rightarrow~~~\bigg(\dfrac 25\bigg)^x=\bigg(\dfrac 25\bigg)^1;~~\boldsymbol{x_2=1}\\\\\boxed{\boldsymbol{x_1=0;~~~x_2=1}}$