Verify if the equation has solution? 5^4x-2*25^x+1=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.