25^x - 6*5^x + 5 = 0

Let us rewrtie 25^x

We know that 25 = 5^2

==> 25^x = 5^x)^2

Then let us rewrtie the equation:

(5^x)^2 - 6*5^x) + 5 = 0

Now let us assume that 5^x = y

==> y^2 - 6y + 5 = 0

Now we will factor:

==> ( y - 5)(y-1) = 0

==> y1= 5 ==> 5^x1 = 5 ==> x1 = 1

==> y2= 1 ==> 5^x2 = 1 ==> x2 = 0

Then we have two solutions:

**x = { 0, 1}**

This is an exponential equtaion and we'll solve it using technique of substitution.

We'll substitute 5^x by the variable t.

25^x=(5^2)^x=(5^x)^2=t^2

We'll re-write the equation in t:

t^2 -6t + 5 = 0

We'll apply the quadratic formula:

t1 = [6+sqrt(36-20)]/2

t1 = (6+4)/2

t1 = 5

t2 = (6-4)/2

t2 = 1

Now, we'll determine x:

5^x = t1

5^x = 5

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

x = 1

5^x = t2

5^x = 1

But 5^0 = 1

5^x = 5^0

x = 0

**The solutions of the equation are: {0 ; 1}.**

To solve

25^x - 6*5^x + 5 = 0.

We notice that 25^x = (5^x)^2.

So we put 5^x = t in the equation 25^x - 6*5^x +5 = 0 and get:

t^2-6t +5 = 0.

t^2-5t -t +5 = 0.

t(t-5) -1(t-5) = 0.

(t-5)(t-1) = 0.

t-5 = 0 , or t-1 = 0.

t-5= 0 gives t= 5 , 5^x = 5 = 5^1.So x = 1.

t-1 = 0 gives t=1, 5^x = 1= 5^0. So x= 0.

Therefore x = 0 or x= 1 are the solutions.