Log2(6) - log2(3) + 2 log 2(root 8) as a single logarithm and then evaluate

please help

log2(6) - log2(3) + 2 log 2(root 8)
= log2(2*3)-log2(3)+2log2(8^(1/2))

remember that a*log(b) = log(b^a), so
= log2(2*3)-log2(3)+log2((8^(1/2))^2)
= log2(2*3)-log2(3) + log2(8)
= log2(2*3)-log2(3) + log2(2^3)
log(a*b) = log(a)+log(b), so
= log2(2) + log2(3)-log2(3) + log2(2^3)
= log2(2)+log2(2^3)
= log2(2^4)
= 4log2(2)
= 4

Its 4. Didnt feel like typing it out but i agree with the second answer.

log2(6/3) + log2(sqrt8^2)
=1+log2(2^3)
=1+3
=4