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

Question

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

Answer 1

 

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\\\\ \[ \boxed{ x = -\frac{2}{5} } \]" alt="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\\\\ \[ \boxed{ x = -\frac{2}{5} } \]" align="absmiddle" class="latex-formula">

 

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