How do you evaluate #e^ln3#?

2 Answers
Feb 12, 2015

Try to write it as a log:
#ln(x)=ln3# which is: #x=e^(ln(3))#
But to have:
#ln(x)=ln3# means that #x=3# or: #e^(ln(3))=3#

Aug 5, 2018

Answer:

#3#

Explanation:

Since base-#e# and natural log (#ln#) are inverses, they cancel, and we're left with

#cancele^(cancel(ln)3)=3#

Hope this helps!