How do you solve #(lnx)^2 -1 = 0#?

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Answer 1 · 2015-12-22T17:41:08

Rearrange then take exponents to find:

#x = e# or #x = 1/e#

Add #1# to both sides of the equation to get:

#(ln x)^2 = 1#

Hence:

#ln x = +-1#

By definition of #ln# we know #e^(ln x) = x#. Hence:

#x = e^(ln x) = e^(+-1)#

That is #x = e# or #x = 1/e#