Question

2 log 4(2x+3)=1+log4x+log4(2x-1)

Answer 1

$2log_4(2x+3)=1+log_4x+log_4(2x-1)\\\\ODZ:\; \; \left \{ {{2x+3>0\; ,\; \; x>0} \atop {2x-1>0}} \right. \; \; \left \{ {{x>0} \atop {x>0,5}} \right. \; \; \Rightarrow \; \; x>0,5\\\\log_4(2x+3)^2=log_4(4x\cdot (2x-1))\\\\4x^2+12x+9=8x^2-4x\\\\4x^2-16x-9=0\; ,\; \; D/4=100\; ,\; \; x_{1,2}=\frac{8\pm 10}{4}\\\\x_1=-0,5\notin ODZ\; ,\; \; x_2=4,5\\\\Otvet:\; \; x=4,5\; .$

Answer 2

Решение во вложении.
Ответ: 4,5