log[tex] log_{6} (x^{2}+x) = log_{2}x \\
\frac{log _{2}x(x+1) }{ log_{2}6} = log_{2}x \\
log_{2}x + log_{2}(x+1)= log_{2}x(log_{2}3+log_{2}2) \\
log_{2}x + log_{2}(x+1)=log_{2}3x+log_{2}x \\
log_{2}(x+1)=log_{2}3x => \\
x+1=3x \\
2x=1\\
x= \frac{1}{2}
\\
\\
c.e.:
x > 0;(1)
x(x+1) > 0;(2)
din (1), (2) => x > 1[/tex]
nu sunt foarte sigura dar cam asa se face
