Question

Решить систему: log(2)(x+y)+2log(4)(x-y)=3 3^(2+log(3)(2x-y)=45 решить не равенство: log(1/4)(2x-5> -1

Answer 1

$\left \{ {{log_2(x+y)+2log_4(x-y)=3} \atop {3^{2+log_3(2x-y)}=45}} \right. \ \ \ \ \ \ \ \ \left \{ {{log_2(x+y)+2log_{2^2}(x-y)=3} \atop {3^2*3^{log_3(2x-y)}=45}} \right. \\ \left \{ {{log_2(x+y)+2*\frac{1}{2}log_{2}(x-y)=3} \atop {9*3^{log_3(2x-y)}=45}} \right. \ \ \ \ \ \left \{ {{log_2(x+y)+log_{2}(x-y)=3} \atop {2x-y}=5}} \right. \\ \left \{ {{log_2((x+y)(x-y))=3} \atop {2x-y}=5}} \right. \ \ \ \ \left \{ {{x^2-y^2=8} \atop {2x-y}=5}} \right. \ \ \ \$
$\left \{ {{x^2-y^2=8} \atop {y=2x-5}} \right. \ \ \ \left \{ {{x^2-(2x-5)^2=8} \atop {y=2x-5}} \right.\\ x^2-4x^2+20x-25-8=0 \\ -3x^2+20x-33=0 \\ 3x^2-20x+33=0 \\ D=400-396=4 \\ x_1=\frac{20+2}{6}= 3\frac{2}{3} \ \ \ \ \ \ \ \ \ \ x_2=\frac{20-2}{6}=3 \\ y_1=2*3\frac{2}{3}-5=2\frac{1}{3} \ \ \ \ \ \ \ y_2=2*3-5=1 \\ \\ (3\frac{2}{3};2\frac{1}{3}) \ \ \ \ \ \ \ (3;1)$

$log_\frac{1}{4}(2x-5)-1 \\ \left \{ {{2x-50} \atop {2x-5<(\frac{1}{4})^{-1}}} \right. \ \ \ \ \left \{ {{2x5} \atop {2x-5<4} \right. \ \ \ \ \left \{ {{x2,5} \atop {2x<9} \right. \ \ \ \ \left \{ {{x2,5} \atop {x<4,5} \right.$
x∈(2,5; 4,5)