How do you solve #log_e x + (log_e x)^2 = 6#?

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Answer 1 · 2018-05-28T13:23:40

#x=e^2#

With #t=ln(x)# we get

#t^2+t-6=0#

solving this equation

#t_1=2#

or

#t_2=-3#(which is not a solution!)

so

#x=e^2#