Question

log (4x-2)=log2-log(2x-1)

Answer 1

$\mathrm{\Rightarrow \log{a}-\log{b}=\log{\bigg(\dfrac{a}{b}\bigg)}}\\\\ \mathrm{\log{(4x-2)}=\log{2}-\log{(2x-1)}\ \to\ \log{(4x-2)}=\log{\bigg(\dfrac{2}{2x-1}\bigg)}\ \to}\\\\ \mathrm{\to\ 4x-2=\dfrac{2}{2x-1}\ \to\ 2x-1=\dfrac{1}{2x-1}\ \to\ (2x-1)(2x-1)=1\ \to}\\\\ \mathrm{\to\ 4x^2-2x-2x+1=1\ \to\ 4x^2-4x+1-1=0\ \to\ 4x(x-1)=0}\\\\ \mathrm{\mathbf{I)}\ 4x=0\ \to\ x=0\ \to\ n\~ao\ conv\acute{e}m\ \ \bigg\|\ \ \mathbf{II)}\ x-1=0\ \to\ \boxed{\mathrm{x=1}}}$