Решите уравнение логарифма

Question 1

Question 2

проверьте условия 2 и 3

Question 3

1) Область определения: x > 0
$log_{0,25}^2(64x)=(log_{1/4}(64)+log_{1/4}(x))^2=(-3+log_{1/4}(x))^2=$
$=9-6log_{1/4}(x)+log_{1/4}^2(x)$
Замена $y=log_{1/4}(x)$
9 - 6y + y^2 + 8y - 17 = 0
y^2 +2y - 8 = 0
(y + 4)(y - 2) = 0
$y1=log_{1/4}(x)=-4$ ; x1 = (1/4)^(-4) = 4^4 = 256
$y2 = log_{1/4}(x) = 2$ ; x2 = (1/4)^2 = 1/16

2) Область определения: x > 0
$lg^2(100x)=(lg(100)+lg(x))^2=(2+lg(x))^2=4+4lg(x)+lg^2(x)$
Замена lg(x) = y
4 + 4y + y^2 + 14y + 76 = 0
y^2 + 18y + 80 = 0
(y + 8)(y + 10) = 0
y1 = lg(x) = -8; x1 = 10^(-8)
y2 = lg(x) = -10; x2 = 10^(-10)

3) Область определения: x > 0
$log_{1/5}^2(625x)=(log_{1/5}(625)+log_{1/5}(x))^2=(-4+log_{1/5}(x))^2=$
$=16-8log_{1/5}(x)+log_{1/5}^2(x)$
$log_{0,04}(x)=log_{1/25}(x)= \frac{lg(x)}{lg(1/25)} = \frac{lg(x)}{2lg(1/5)} =0,5*log_{1/5}(x)$
Замена $log_{1/5}(x)=y$
16 - 8y + y^2 +14*0,5y = 8
y^2 -8y + 7y + 16 - 8 = 0
y^2 - y + 8 = 0
Это квадратное уравнение решений не имеет.

mefody66_zn · Answer 1 · 2018-05-16T02:46:47+0000

1) Область определения: x > 0
$log_{0,25}^2(64x)=(log_{1/4}(64)+log_{1/4}(x))^2=(-3+log_{1/4}(x))^2=$
$=9-6log_{1/4}(x)+log_{1/4}^2(x)$
Замена $y=log_{1/4}(x)$
9 - 6y + y^2 + 8y - 17 = 0
y^2 +2y - 8 = 0
(y + 4)(y - 2) = 0
$y1=log_{1/4}(x)=-4$ ; x1 = (1/4)^(-4) = 4^4 = 256
$y2 = log_{1/4}(x) = 2$ ; x2 = (1/4)^2 = 1/16

2) Область определения: x > 0
$lg^2(100x)=(lg(100)+lg(x))^2=(2+lg(x))^2=4+4lg(x)+lg^2(x)$
Замена lg(x) = y
4 + 4y + y^2 + 14y + 76 = 0
y^2 + 18y + 80 = 0
(y + 8)(y + 10) = 0
y1 = lg(x) = -8; x1 = 10^(-8)
y2 = lg(x) = -10; x2 = 10^(-10)

3) Область определения: x > 0
$log_{1/5}^2(625x)=(log_{1/5}(625)+log_{1/5}(x))^2=(-4+log_{1/5}(x))^2=$
$=16-8log_{1/5}(x)+log_{1/5}^2(x)$
$log_{0,04}(x)=log_{1/25}(x)= \frac{lg(x)}{lg(1/25)} = \frac{lg(x)}{2lg(1/5)} =0,5*log_{1/5}(x)$
Замена $log_{1/5}(x)=y$
16 - 8y + y^2 +14*0,5y = 8
y^2 -8y + 7y + 16 - 8 = 0
y^2 - y + 8 = 0
Это квадратное уравнение решений не имеет.