How do you solve #ln (1/e^5)=x#?

1 Answer
Jul 22, 2016

Answer:

The soln. is # : x=-5

Explanation:

#ln(1/e^5)=x#

We know that, #1/e^a=e^(-a), and, lne^b=b#

#:. ln(e^(-5))=ln (e^x))............(1)#

Now, #ln# fun. is #1-1#. Hence, #(1) rArre^(-5)=e^x#.

Again, fun. #e^x# is #1-1#, so, #x=-5# is the soln.