The required value of x is 1.

$$\rm 4^{2x - 1} = 2^{x+1}$$

By using the law of indices,

$$\rm \left ( 2^2 \right )^{2x - 1} = 2^{x+1}$$

By using the identity that $\rm (a^m)^n = a^{mn}$, we get,

$$\rm 2^{2(2x-1)} = 2^{x+1}$$

Since the bases of both Left Hand and Right Hand sides of the equation are same, we equate their exponents, we get,

$$\rm 2(2x - 1) = x+ 1$$

$$\rm or, 4x - 2 = x + 1$$

$$\rm or, 4x - x = 2 + 1$$

$$\rm or, 3x = 3$$

$$\rm or, \frac{3x}{3} = \frac{3}{3}$$

$$\rm \therefore x = 1$$