Given,

$$\rm 4^x - 5 \cdot 2^x + 4 = 0$$

$$\rm (2^2)^x - 5 \cdot 2^x + 4 = 0$$

$$\rm (2^x)^2 - 5 \cdot (2^x) + 4 = 0$$

Let $\rm a = 2^x$

$$\rm (a)^2 - 5 (a) + 4 = 0$$

Now, we have a quadratic equation in a. We solve the quadratic equation by factorization.

$$\rm a^2 - (1 + 4) a + 4 = 0$$

$$\rm or, a^2 - a - 4a + 4 = 0$$

$$\rm or, a ( a - 1) - 4 ( a - 1) = 0$$

$$\rm or, (a - 4)(a - 1) = 0$$

Either

$$\rm (a - 4) = 0$

$$\rm or, a = 4$$

$$\rm or, 2^x = 4$$

$$\rm or, 2^x = 2^2$$

The base in the left-hand and the right-hand sides of the equations are the same, so we equate their powers.

$$\rm \therefore x = 2$$

Or

$$\rm (a - 1) = 0$$

$$ \rm or, a = 1$$

$$\rm or, a = 2^0$$

$$\rm or, 2^x = 2^0$$

The base in the left-hand and the right-hand sides of the equations are the same, so we equate their powers.

$$\rm \therefore x = 0$$

Hence, the required values of x are x = {0,2}.