How do you evaluate #ln (1/e)#?

1 Answer
Jun 10, 2016

Answer:

It is #-1#.

Explanation:

We apply the properties of the logarithm:

#ln(1/e)=ln(e^(-1))#

the first property is that the exponent "exit" and multiply the log

#ln(e^-1)=-ln(e)#

the second property is that the logarithm of the base is 1. The base of the natural logarithm is #e# then

#-ln(e)=-1#.

In conclusion

#ln(1/e)=-1#.