What's an intuitive way to compute summation of this series?

Question

What's an intuitive way to compute $$\log(1)+\log (2)+\log (3)+\cdots+\log (n-1)+\log (n)$$ or

for $n>a$ $$\log(a)+\log (a+1)+\log (a+2)+\cdots+\log (n-1)+\log (n) $$

I know the answer for general case is $$\log \left(\frac{n!}{(a-1)!}\right)$$

Answer 1

Note that

$$ \log(1)+\log (2)+\log (3)+....\log (n-1)+\log (n)=\log (n!) $$

and

$$ \log(a)+\log (a +1)+\log (a + 2)+....\log (n-1)+\log (n)= \log \prod_{i=a}^{n}i $$

Note: You need the property

$$ \log(ab) = \log(a)+\log(b). $$