log4(x-3)+log4(x+2)=log4(14),
log4[(x-3)(x+2)]=log4(14),
(x-3)(x+2)=14,
x2-x-6-14=0,
x2-x-20=(x-5)(x+4)⇒x=5 or x=-4, but we need to check the validity of the original equation.
When x=-4, x-3=-7 and the log of a negative number cannot be evaluated. Therefore x=5 is the solution:
log4(2)+log4(7)=log4(14) checks out OK.