How to solve #3sqrt(lnx)+lnx=4#?

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Answer 1 · 2018-05-30T17:14:19

#x=e#

Substituting

#sqrt(ln(x))=t#
so we get
#t^2+3t-4=0#

With the Solutions

#t_1=1#

#t_2=-4<0# (which isn't a solution)

so we have
#sqrt(ln(x))=1#
#ln(x)=1#

so #x=e#