Question

1) logx(2x+3)< 2 2) log4(2x-1)< = log4(x-3) 3) lg^2(10x)-lgx > =3 4) (lg (sqrt(x+1)+1)) / lg^3(sqrt x-40) = 3 как

Answer 1

$1) \ log_x(2x+3)0; \ x-1,5; \ x0, x \neq 1; \ x \in (0;1) \cup (1;+\infty) \\ a) \ 0 x^{2} \\ x^{2} -2x-3 2x+3< x^{2} \\ x^{2} -2x-30 \\ x \in (- \infty;-1)\cup (3;+\infty) \\ x\in (3;+\infty) \\ Answer: x \in (0;1) \cup (3;+\infty)$
$2) \ log_4(2x-1) \leq log_4(x-3) \\ ODZ: \left \{ {{2x-10} \atop {x-30}} \right. = \left \{ {{x0,5} \atop {x3}} \right.= x3 \\ log_4 \frac{2x-1}{x-3} \leq 0 \\ \frac{2x-1}{x-3} \leq 1 \\ \frac{2x-1-x+3}{x-3} \leq 0 \\ \frac{x+2}{x-3} \leq 0 = x \in [-2;3) ; x3 \\ Answer: x \in \varnothing$
$3) \ lg^210x-lgx \geq 3 \\ ODZ: x0 \\ (1+lgx)^2-lgx-3 \geq 0 \\ lgx=y \\ (1+y)^2-y-3 \geq 0 \\ y^2+2y+1-y-3 \geq 0 \\ y^2+y-2 \geq 0 \\ D=9=3^2 \\ y_1=-2;y_2=1 \\ lgx=-2 = x_1=10^{-2}=0,01 \\ lgx=1 = x_2=10 \\ x \in(- \infty;0,01] \cup [10; +\infty);x0 \\ Answer: x \in (0;0,01]\cup [10; +\infty)$
$4) \ \frac{lg( \sqrt{x+1}+1 )}{lg(\sqrt{x} -40)^3}=3 \\ ODZ: \sqrt{x+1}+10 \\ \sqrt{x+1}-1 \\ x+1 \geq 0=x \geq -1 \\ \sqrt{x} -400 \\ x0; \sqrt{x} 40=x1600 \\ lg( \sqrt{x+1}+1 )=3*1/3* lg(\sqrt{x} -40) \\ lg(\sqrt{x} -40)= lg( \sqrt{x+1}+1 ) \\ \sqrt{x} -40=\sqrt{x+1}+1 = \sqrt{x} - \sqrt{x+1}=41 \\ x-2 \sqrt{x(x+1)} +x+1=1681 = 2 \sqrt{x(x+1)}=2x-1680 \\ \sqrt{x(x+1)}=x-840 \\ x^{2} +x= x^{2} -1680x+705600=1681x=705600 \\ x\approx 418; x1600 = Answer: x \in \varnothing$

Answer 2

1)2x+3>0⇒2x>-3⇒x>-1,5 U x>0 U x≠1⇒x∈(0;1) U (1;∞)
a)x∈(0;1)
2x+3>x²⇒x²-2x-3<0
x1+x2=2 U x1*x2=-3⇒x1=-1 U x2=3
-1<x<3 U x∈(0;1)⇒x∈(0;1)
b)x∈(1;∞)
2x+3<x²⇒x²-2x-3>0
x1+x2=2 U x1*x2=-3⇒x1=-1 U x2=3
x<-1 U x>3 U x∈(1;∞)⇒x∈(3;∞)ответ x∈(0;1) U (3;∞)
2)2x-1>0⇒2x>1⇒x>1/2
x-3>0⇒x>3
2x-1≤x-3
2x-x≤-3+1
x≤-2
нет решения
3)x>0
(1+lgx)²-lgx≥3
lgx=a
(1+a)²-a-3≥0
1+2a+a²-a-3≥0
a²+a-2≥0
a1+a2=-1 U a1*a2=-2⇒a1=-2 U a2=1
a≤-2⇒lgx≤-2⇒x≤0,01 U a≥1⇒lgx≥1⇒x≥10
ответ x∈(0;0,01] U [10;∞)