Question 617926
log[6](x-2)+log[6](x+3)=1
log[6][(x-2)(x+3)]=1
(x-2)(x+3) = 6^1
x^2+x-6 = 6
x^2+x-12 = 0
(x+4)(x-3) = 0
x = {-4, 3}
Throw out the negative solution (extraneous) leaving:
x = 3