Question 11673
first, pedantic...it is ln, not In.


OK, here goes:


ln(ln(e)) + ln(e^2)
ln(ln(e)) + 2ln(e)

Since the rules of logs say a "power" can be brough to the front of the log.


Taking "natural log" and the number "e" are opposites, just like + and - are opposites. ln(e) is 1.


so we have ln(1) + 2


Any log(1) is zero, so we have 2 as the answer.


jon.