Question

ЛОГАРИФМИЧЕСКАЯ СИСТЕМА УРАВНЕНИЙ, С ОДЗ

Answer 1

$\left\{\begin{array}{l}lgx-5^{y}+2y=3\\2y\cdot 5^{y}+5^{y}\cdot lgx=4\end{array}\right\; \; \; \; ODZ:\; x0\\\\\\\left\{\begin{array}{l}lgx=3-2y+5^{y}\\lgx=\frac{4-2y\cdot 5^{y}}{5^{y}}\end{array}\right\; \; \; \Rightarrow \; \; \; 3-2y+5^{y}=\frac{4-2y\cdot 5^{y}}{5^{y}}\; \Big |\cdot 5^{y}\; ,\; \; (5^{y}0)\\\\\\5^{y}\cdot (3-2y+5^{y})=4-2y\cdot 5^{y}\\\\(3-2y)\cdot 5^{y}+(5^{y})^2-4+2y\cdot 5^{y}=0\\\\(5^{y})^2+3\cdot 5^{y}-4=0$

$t=5^{y}0:\; \; t^2+3t-4=0\; \; ,\; \; t_1=1\; ,\; t_2=-4\; \; (teorema\; Vieta)\\\\5^{y}=1\; \; \to \; \; 5^{y}=5^0\; \; ,\; \; \underline {y=0}\\\\lgx=3-2y+5^{y}=3-2\cdot 0+1=4\; \; ,\; \; lgx=4\; \; \to \; \; \underline {x=10^4}\\\\Otvet:\; \; (10000\, ,\, 0)\; .$

Answer 2

Объяснение:

Одз х>0