How do you evaluate #e^ln3#?

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Answer 1

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#

Answer 2

Aug 5, 2018

#3#

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

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

Hope this helps!