If #ln x - ln (1/x) = 2#, then how do you find #x#?

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Answer 1 · 2016-10-25T17:49:03

#x = e#

We use the rule #lna - lnb = ln(a/b)# to start the solving process.

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

#ln(x^2) = 2#

If #lna = b#, #e^b= a#

#x^2 = e^2#

#x = +-sqrt(e^2)#

#x = +-e#

However, the negative answer is not possible as it renders the original equation undefined.

Hopefully this helps!