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

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Answer 1 · 2015-07-24T19:22:53

#x in {-e^2, e^2}#

#lnx^2=4#

#=>x^2=e^4#

#=>x^2-e^4=0#

Factorize,

#=>(x-e^2)(x+e^2)=0#

There are two solution,

#=>x-e^2=0 =>x=e^2#

And,

#=>x+e^2=0=>x=-e^2#