Question

1)решите неравенства: а) log4 (x-2)< 2; б)log0,5 (4x-7) 2) решите систему уравнений : а) 3^y*9^x=81, lg(x+y)^2-lg x=2 lg3; .

Answer 1

$1)\quad log_4(x-2)\ \textless \ 2\; ,\; \; \ ;ODZ:\; x\ \textgreater \ 2\\\\log_4(x-2)\ \textless \ log_44^2\\\\x-2\ \textless \ 16\\\\x\ \textless \ 18\\\\x\in (2,18)\\\\2)\quad \left \{ {{3^{y}\cdot 9^{x}=81} \atop {lg(x+y)^2-lgx=2lg3}} \right. \; \left \{ {{3^{y+2x}=3^4} \atop {lg\frac{(x+y)^2}{x}=lg3^2}} \right. \; \left \{ {{y+2x=4} \atop {\frac{x^2+2xy+y^2}{x}=9} \right. \\\\ \left \{ {{y+2x=4} \atop {x^2+2xy+y^2=9x}} \right. \; \left \{ {{y=4-2x} \atop {x^2+2x(4-2x)+(4-2x)^2=9x}} \right. \\\\x^2+8x-4x^2+16-16x+4x^2-9x=0\\\\x^2-17x+16=0\\\\D=17^2-4\cdot 16=225$

$x_1=\frac{17-15}{2}=1\; ,\; \; x_2=16\\\\y_1=4-2=2\; ,\; \; y_2=4-32=-28\\\\ODZ:\; x+y \geq 0\; ,\; x \geq 0\\\\x \geq -y\; ,\; x \geq 0\\\\x+y=1+2=3 \geq 0\\\\x+y=16-28=-12\ \textless \ 0\\\\Otvet:\; \; (1,2)\; .$

$3)\quad log_{1/7}(4x+1)\ \textless \ -2\; ,\quad ODZ:\; x\ \textgreater \ -\frac{1}{4}\\\\log_{1/7}(4x+1)\ \textless \ log_{1/7}(1/7)^{-2}\\\\4x+1\ \textgreater \ (1/7)^{-2}\\\\4x+1\ \textgreater \ 49\\\\4x\ \textgreater \ 48\\\\x\ \textgreater \ 12\\\\x\in (12,+\infty )$

$4)\quad log_{0,5}(4x-7)\ \textless \ log_{0,5}(x+2)\; ,\quad ODZ:\; x\ \textgreater \ \frac{7}{4}\; ,\; x\ \textgreater \ 1,75\\\\4x-7\ \textgreater \ x+2\\\\3x\ \textgreater \ 9\\\\x\ \textgreater \ 3\\\\x\in (3,+\infty )$