log_b założenia a>0 , a \ne 1
log_{2x+1} 5 = 2
2x+1>0 \\\\2x>-1\\\\ x>-\frac{1}{2}
D: \ x\in (-\frac{1}{2};+\infty)
(2x+1)^2=5\\\\ 4x^2+4x+1-5=0\\\\4x^2+4x-4=0\ |:4\\\\x^2+x-1=0\\\\\Delta=b^2-4ac=1-4\cdot1\cdot (-1)=5
x_1=\frac{-1-\sqrt5}{2}\approx-1,61\not\in D
x_2=\frac{-1+\sqrt5}{2}=\frac{\sqrt5-1}{2}
\approx 0,61\in D
x=\frac{\sqrt5-1}{2}