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How do you solve #Ln x^2=4#?

1 Answer

Aug 4, 2015

I found: #x=e^2=7.389#

Explanation:

You can use a property of the logs and write:
#lnx^2=2lnx#
so that you have:
#2lnx=4#
#lnx=4/2=2#
so that #x=e^2#

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