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

1 Answer
Ratnaker Mehta
Jul 22, 2016

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.

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