Solve the equation #2(lnx)^2-5lnx+2=0#?

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Answer 1 · 2018-01-13T13:42:12

#x=sqrte# or #e^2#

#(lnx)^2=lnx*lnx=ln(x^(lnx))#. But it is complicated this way.

Just take #lnx=u# and equation becomes

#2u^2-5u+2=0#

or #(2u-1)(u-2)=0#

Hence #u=1/2# or #u=2#

therefore either #lnx=1/2# i.e. #x=e^(1/2)=sqrte#

or #lnx=2# i.e. #x=e^2#