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##