Question

log2(2x-1)+log2(x+3)=2

Answer 1

Ответ: х=1.

Объяснение:

$log_2(2x-1)+log_2(x+3)=2$.

ОДЗ:   $\displaystyle\\\left \{ {{2x-1 > 0} \atop {x+3 > 0}} \right. \ \ \ \ \left \{ {{2x > 1\ |:2} \atop {x > -3}} \right. \ \ \ \ \left \{ {{x > 0,5} \atop {x > -3}} \right.\ \ \ \ \Rightarrow\ \ \ \ x\in(0,5;+\infty).$

$\displaystyle\\log_2(2x-1)(x+3)=log_24\\\\(2x-1)(x+3)=4\\\\2x^2-x+6x-3=4\\\\2x^2+5x-7=0\\\\2x^2-2x+7x-7=0\\\\2x*(x-1)+7*(x-1)=0\\\\(x-1)*(2x+7)=0$

x-1=0

x₁=1 ∈ОДЗ.

2х+7=0

2х=-7 |:2

x=-3,5 ∉ОДЗ.