Question

logx (x+1) = 2, multumesc!

Answer 1

$log_{x} (x+1)=2\\=> x+1=x^{2} \\=> -x^{2} +x+1=0\\x^{2} -x-1=0\\=> x_{12} =\frac{-(-1)+\sqrt{1+4} }{2} =\frac{1+\sqrt{5} }{2}$