How do you solve #e ^(lnx-3lnx) = (5/2)#?

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Answer 1 · 2016-04-17T10:09:14

#x=sqrt(2/5)#

Note that #x# cannot be negative.

Taking log of both sides to base #e# we get

#lnx-3lnx=ln(5/2)# now simplifying LHS

#lnx-3lnx=lnx-lnx^3=ln(x/x^3)=ln(1/x^2)=ln(5/2)#

Hence #1/x^2=5/2# or #x^2=2/5#

Hence #x=+-sqrt(2/5)# but x cannot be negative.

Hence #x=sqrt(2/5)#.