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$ln(a) + ln(b) = ln(a*b)$ Therefore we can evaluate $n$ as being $12*11*10*9*8*7*6*5*4*3*2*1 =12! =479001600$ .

Since the sum of logarithms of numbers is equal to the logarithm of their product, the sum of the given logs = log(12!). Ed Gray

$\ln 1 + \ln 2 + \ln 3 + \cdots + \ln 12 = \ln (1\times 2 \times 3 \times \cdots \times 12) = \ln(12!)$ $\implies n = 12! = \boxed{479001600}$