You need to convert the given exponential equation a quadratic equation, such that:

`5^x = y => 5^(4x) = (5^x)^4 => (5^x)^4 = y^4`

Changing the variable yields:

`y^4 - 2y^2 + 1 = 0 => (y^2 - 1)^2 = 0 => y^2 - 1 = 0 => y^2 = 1 => y_(1,2) = +-1`

Substituting back `y` for `5^x` yields:

`5^x = y_1 => 5^x = 1 => ln 5^x = ln 1 => x*ln 5 = 0 => x = 0`

`5^x = y_2 => 5^x = -1` invalid

**Hence, evaluating the solution to exponential equation yields `x = 0.` **

The given exponential equation requires substitution technique to compute it's roots.

Now, we notice that 25 = 5^2

We'll re-write the equation as:

5^4x - 2*5^2x + 1 = 0

It is a bi-quadratic equation:

We'll substitute 5^2x by another variable.

5^2x = t

We'll square raise both sides:

5^4x =t^2

We'll re-write the equation, having "t" as variable.

t^2 - 2t + 1 = 0

The equation above is the result of expanding the square:

(t-1)^2 = 0

t1 = t2 = 1

But 5^2x = t1.

5^2x = 1

We'll write 1 as a power of 5:

5^2x = 5^0

Since the bases are matching, we'll apply the one to one property:

2x = 0

We'll divide by 2:

x = 0.

The equation has a solution and it is x = 0.