Question 283616
Log6 (x+3) + Log6 (x-2) = 1



Log6 [(x+3)(x-2)] = 1



(x+3)(x-2) = 6^1


(x+3)(x-2) = 6



x^2+x-6 = 6


x^2+x-12=0


(x+4)(x-3)=0


x+4=0 or x-3=0


x=-4 or x=3


x=-4 is extraneous


Answer: Only solution is x=3