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

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Answer 1 · 2016-06-10T11:49:52

It is #-1#.

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#.