How do you solve #log_3 (x + 3) + log_3 (x + 5) = 1#?
log(base3)((x+3)(x+5))=1 write in exponential form
x=-6 or x=-2
x=-6 is extraneous . An extraneous solution is root of transformed but it is not a root of original equation.
so x=-2 is the solution.