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[(e^x) + (e^-x)]/2 = 1

First ley us multiply by 2:

===> (e^x) + e^-x =  2

Now let us assume that:

y = e^x  ==> e^-x = 1/y

==> y + 1/y = 2

Multiply by y:

==> y^2 + 1 = 2y

==> y^2 - 2y + 1 = 0

==> (y-1)^2 = 0

==> y = 1

==> e^x = 1

==> x= 0

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