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

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### 3 Answers

[(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**

This is an exponential equation and we'll solve it using the substitution technique.

First, we'll note that we have a negative power:

e^-x = 1/e^x

Now, we'll note e^x = t

We'll re-write the equation in t:

(t + 1/t) / 2 = 1

We'll cross multiply:

t + 1/t = 2*1

t + 1/t = 2

We'll multiply by t;

t^2 + 1 = 2t

We'll subtract 2t both sides:

t^2 - 2t + 1 = 0

But the expression is resulted after expanding the square

(t-1)^2 = 0

We'll put t-1 = 0

t = 1

But e^x = t, so e^x = 1 (1)

We could write 1 = e^0.

We'll re-write (1):

e^x = e^0

x = 0

**The solution of the equation is x = 0.**

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

Solution:

In fact for any number a, (a^x+a^-x)/2 = 1 thw obvious answer is x= 0 by substotution x =0 in the equation (a^x+a^-x)/2 = (a^0+a^0)/2 = 1.

So it should hold good for a = e.

Now let us go by procedure:

(e^x+e^-x)/2 = 2. Multiply by 2e^x.

e^2x +1 = 2e^x.

(e^x)^2 - 2e^x+1 = 0

{e^x-1}^2 = 0

e^x -1 = 0

e^x = 1 . But 1 = e^0

e^x = e^0.

Bases are same on both sides. So exponents must be equal:

x = 0.