How do you solve #lnx=2#?

1 Answer
Jul 18, 2016

Answer:

#x=7.3891#

Explanation:

If #a^n=b#, we have #log_ab=n#, where #a# is called base. When base is #a=e#, we have Napier's logarithm and we need not write base and we just write it as #lnb=n# which is equivalent to #b=e^n#.

Hence, #lnx=2#, can be written as #x=e^2# and as #e^2=7.3891#, #x=7.3891#