Question

`решите логарифмы пожалуйста

Answer 1

$log_{2} (x^{2} +3x-4)>1+log_{2} (4x-5)$

$log_{2} (x^{2} +3x-4)>log_{2}2+log_{2} (4x-5)$

$log_{2} (x^{2} +3x-4)>log_{2}(2*(4x-5))$

x²+3x-4>8x-10

x²-5x+6>0

x²-5x+6=0

D = (-5)² - 4*6 = 25 - 24 = 1

$x_{1} = \frac{5-1}{2} = 2$

$x_{2} = \frac{5+1}{2} = 3$

x∈(-∞;2)∪(3;+∞)

ОДЗ:

$\left \{ {{x^{2}+3x-4>0 } \atop {4x-5>0}} \right.$

x²+3x-4=0

D = 3² - 4*(-4) = 9 + 16 = 25 = 5²

$x_{1} = \frac{-3+5}{2} = 1$

$x_{2} = \frac{-3-5}{2} = -4$

x∈(-∞;-4)∪(1;+∞)

4x - 5 = 0

4x = 5

x = 5/4

$\left \{ {{x>1,25} \atop {\\\left[\begin{array}{ccc}x<-4\\x>1\\\end{array}}} \right.$

x∈(1,25;+∞)

Ответ: x∈(1,25;2)∪(3;+∞)