Given,

$\rm 3^{2x + 1} = 9^{2x - 1}$

$\rm or, 3^{2x + 1} = (3^2)^{2x  -1}$

Using the law of indices: $\rm (a^b)^c = a^{bc}$ 

$\rm or, 3^{2x + 1} = 3^{2(2x  -1)}$

The base of the terms in the left-hand and right-hand sides of the equations are the same; hence, we equate their exponents.

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

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

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

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

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

$\rm or, 3 = 2x$

$\rm \therefore x = \frac{3}{2}$

Hence, the required value of x = $\frac{3}{2}$ .