# Verify if the equation has solution? 5^4x-2*25^x+1=0

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To find the solution of 5^4x-2*25^x+1 = 0.

Solution:

We know 25^x = 5^2x .

5^4x = (5^2x)^2 = (5^2x)^2.

We put 5^2x = t in the given equation :

t^2 - 2t+1 = 0.

Therefore (t-1)^2 = 0.

(t-1) = 0.

t = 1.

Therefore 5^2x = 1 = 5^0

5^2x = 5^0.

2x= 0.

So x= 0.

Therefore 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.**