# Find x for the identity is true 1+12^6x = 2*12^3x

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1 + 12^6x = 2*12^3x

Let us rewrite:

(12^3x)^2 - 2*12^3x + 1 = 0

Let y = 12^3x

==> y^2 - 2y + 1 = 0

==> (y-1)^2 = 0

==> y= 1

==> y= 12^3x = 1

==> 3x = 0

**==> x = 0**

1+12^6x=2*12^3x.

Put 12^3x = t in the given equation. Then 12^6x = (12^3x)^2 = t^2. So the equation becomes"

1+t^2= 2t

1+t^2 - 2t = o

t^2-2t+1 = 0

(t-1)^2 = 0

So t-1 = 0. Or 12^3x - 1 = 0

a2^3x =1 = 12^0.

3x = 0.

x = 0.

**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.

12^6x - 2*12^3x + 1 = 0

It is a bi-quadratic equation:

We'll substitute 12^3x by another variable.

12^3x = t

We'll square raise both sides:

12^6x = t^2

We'll re-write the equtaion, 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 12^3x = t1.

12^3x = 1

We'll write 1 as a power of 12:

12^3x= 12^0

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

3x = 0

We'll divide by 3 both sides:

x = 0.

**The solution of the equation is x = 0.**