# Solve for x ,5^4x = 2*25^x – 1

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**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.**

5^(4x) = 2*25^x -1.

5^(4x) = (25^2)^x = (25^x)^2.

So we put 25^x = t and rewrite the given equation:

t^2 = 2t -1

t^2-2t +1 = 0

(t-1)^2 = 0

Therefore t -1 = 0

t = 1. Or

25^x = 1. But 1 = 25^0 also.

25^x = 25^0.

x = 0.

We can rewrite the given equation using the relation a^(b*c)= (a^b)^c

Now 25 = 5^2, so we have:

5^4x = 2*25^x – 1

=> 5^2^2x = 2*25^x -1

=> 5^2x^2 = 2*5^2x -1

Let Y = 5^2x

=> Y^2 = 2Y – 1

=> Y^2 – 2Y+ 1 = 0

=> (Y-1)^2 = 0

taking the square root of both the sides

=> Y-1 = 0

=> Y =1

As Y = 5^2x, so 5^2x = 1

Now only if 2x= 0 can 5^2x be equal to 0 as any number raised to the power 0 is 1.

**So the required solution is x=0.**