Question

2logx-2log (x+1)=0
urgente

Answer 1

${2log(x)-2log(x+1)=0}\\\\{2log(x)=2log(x+1)}\\\\{log(x)^2=log(x+1)^2}\\\\{x^2=(x+1)^2}\\\\{x^2=x^2+2x+1}\\\\{x^2-x^2-2x-1=0}\\\\{-2x-1=0}\\\\{-2x=1}\\\\{x=-1/2}\\\\{\mathbf{salu2.\ :)}}\\{\mathbf{Wellington}}$