How do you solve # lnx-ln(1/x)=2#?

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Answer 1 · 2016-07-19T17:44:24

#x = ± e#

We can apply a property of logarithms, namely the property

#ln(a) - ln(b) = ln(a/b)#

Thus, we have

#ln(x) - ln(1/x) = 2#

#ln(x/(1/x)) = 2#

#ln(x^2) = 2#

#x^2 = e^2#

#x = ± e#