Solve for x ,5^4x = 2*25^x – 1

This is an exponential equation that
requires substitution technique.

First,
we'll move all terms to one side, changing the sign of the terms
moved.

5^4x - 2*25^x + 1 =
0

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 =
a

We'll square raise both
sides:

5^4x  =a^2

 We'll
re-write the equtaion, having "a" as variable.

a^2 - 2a + 1
= 0

The equation above is the result of expanding the
square:

(a-1)^2 = 0

a1 = a2 =
1

But 5^2x = a1.

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 solution of the equation is x =
0.