Thid is a another log problem solve 2+Logbase2[x+2]= Logbase2[x]+ Logbase2[x-3]?

Question

solve 2+Logbase2[x+2]= Logbase2[x]+ Logbase2[x-3]

Answer 1

4(x+2) = x(x-3)
x^2-7x-8 = (x-8)(x+1) = 0
x = 8, -1
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Ideas: 2 = Logbase2[4]

Answer 2

2+ log(x+2) = log(x) +log(x-3)
2 = log(x) +log(x-3)-log(x+2)
2 = log [(x^2-3x)/(x+2)]
2^2 = (x^2-3x)/(x+2)
4x +8 = x^2-3x
x^2-7x-8 = 0
(x-8)(x+1) = 0
x = 8
x = -1 is rejected beacause it is not in the domain of log(x)