What is the natural log of zero?

This is a tricky question because you do not have a unique answer...I mean, you do not have an answer such as: "the result is 3".
The problem here rests in the definition of log:
#log_ax=b -> x=a^b#
so basically with the log you are looking for a certain exponent that when you rise the base to it gives you the integrand.

Now, in your case you have:
#log_e0=ln0=b#
where #ln# is the way to indicate the natural log or log in base #e#.

But how do you find the right #b# value such that #e^b=0#????

Actually it doesn't work...you cannot find it...you cannot rise to the power of a number and get zero!
If you try with a positive #b# it doesn't work (it gets bigger and not zero); for #b=0# it is even worse because you get #e^0=1#!
One thing you can do is to manipulate it to get as near as possible to zero...
if you take a negative exponent you can get almost there:
if #b# is VERY big (negatively) you get very near to zero:

for example: #e^-100=1/e^100=3.72xx10^-44#!!!!

basically if #b->-oo# then #x=e^b->0#
So I would say that #ln0->-oo# using "tends to" instead of “equal to”.