We have to evaluate 4^(x^2-4x) = 1/64.

Now we know that 1/64 = 64^-1 = (4^3)^-1 = 4^(-3)

Now 4^(x^2-4x) = 1/64

=> 4^(x^2-4x) = 4^(-3)

=> (x^2-4x) = -3

=> x^2 - 4x + 3 =0

=> x^2 - 3x -x +3 =0

=> x(x-3) - 1(x-3) =0

=> (x-1)(x-3) =0

This gives x =1 and x=3.

**Therefore the required values of x are x =1 and x=3**

This is an exponential equation.

For the beginning, we'll write 1/64 = 1/4^3

We'll apply the property of negative power:

1/4^3 = 4^(-3)

Now, we'll re-write the equation:

4^(x^2-4x) = 1/64

4^(x^2-4x) = 4^(-3)

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

x^2-4x = -3

We'll add 3 both sides:

x^2-4x+3 = 0

According to the rule, the quadratic equation could be written as:

x^2 - Sx + P = 0

From Viete's relations, we'll get:

S = x1 + x2

x1 + x2 = 4

x1*x2 = 3

**x1 = 1**

**x2 = 3**

**The solutions of the equation are :{1 ; 3}.**