How do you solve log ( x +1 ) + log 7 = log 14 – log ( 2-x ) ?

As ##logA+logB=logAB##
& ##logA-logB=log(A/B)##
Therefore ##log(x+1)+log7=log14-log(2-x)## can be written as
##log7(x+1)=log(14/(2-x))##
both sides are in the form of single term logs to the same base so now we can drop the log from both sides
##7(x+1)=14/(2-x)##
##(x+1)(2-x)=14/7##
##-x^2+x+2=2##
##x^2-x=0##
##x(x-1)=0##
##x=0## OR ##x=1##