Question

Найдите корень уравнения: log2 (4-x) = log2(1-3x) + 1

Answer 1

 

$log_2(4-x) = log_2(1 - 3x) + 1\\ 4 - x > 0, x < 4\\ 1 - 3x > 0, -3x > -1, x < frac{1}{3}\\ log_2(4-x) = log_2(1 - 3x) + log_22\\ log_2(4-x) = log_22(1 - 3x)\\ log_2(4-x) = log_2(2 - 6x)\\ 4 - x = 2 - 6x\\ -x + 6x = 2 - 4\\ 5x = -2\\ [ boxed{ x = -frac{2}{5} } ]$

 

Answer 2

log2 (4-x) = log2(1-3x) + 1               4-х >0  и 1-3х>0

log₂ (4-x) = log₂(1-3x) +  log₂2  

log₂ (4-x) = log₂(1-3x) ·2)

4- х = 2( 1-3х)

4-х = 2 - 6х

-х +6х = 2- 4

5х = -2

х= -0,4

-0,4 удовлетворяет условиям: 4-х >0  и 1-3х>0

Ответ : -0,4