# Solve the equation 4^4x-2*16^x+1=0

*print*Print*list*Cite

### 2 Answers

**This is an exponential equation that requires substitution technique.**

Now, we notice that 16 = 4^2

We'll re-write the equation as:

4^4x - 2*4^2x + 1 = 0

It is a bi-quadratic equation:

We'll substitute 4^2x by another variable.

4^2x = a

We'll square raise both sides:

4^4x =a^2

We'll re-write the equation, 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 4^2x = a1.

4^2x = 1

We'll write 1 as a power of 4:

4^2x = 4^0

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.**

4^4x-2*16^x+1 = 0

To solve the eqation we put 16^x = y.

Therefore 4^4x = 16^2x = (16^x)^2 = y^2.

With this transformationm the give equation becomes:

y^2-2y+1 = 0

(y-1)^2 = 0

Therefore y - 1 = 0.

y = 1.

y = 16^x = 1.

16^x = 1= 16^0.

x = 0.