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;
Спасибо.

Answer 1

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

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

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

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