How do you solve #ln e^(sqrt{2})#?

1 Answer
Daniel L.
Oct 18, 2016

The value of this expression is #sqrt(2)#

Explanation:

This question ca be answered using the definition of logarythm (here natural #ln#))

According to a definition:

#log_a b=c iff a^c=b#

So we are looking for a number to which the base has to be risen to get the value of #b#.

In the given example we are looking for a number to which we have to raise #e# to get the number #e^sqrt(2)#. From the form of the number #e^sqrt(2)# we can see that the answer is #sqrt(2)#.

Related questions
See all questions in Common Logs
Impact of this question
2522 views around the world
You can reuse this answer
Creative Commons License