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

[(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