Given

$\rm 9^{x} = 4\cdot 3^{x + 1} - 27$

By the law of indices, $\rm a^{m + n} = a^{m} \cdot a^{n}$, we get,

$\rm or, (3^{2})^{x}  = 4 \cdot 3^{x} \cdot 3^{1} - 27$

By the law of indices, $\rm ( a^{m} ) ^{n} = a^{mn}$, we get,

$\rm or, 3^{2x} = 12 \cdot 3^{x} - 27$

By the law of indices, $\rm a^{mn} = ( a^{m} ) ^{n}$, we get,

$\rm or, (3^{x})^{2} = 12 \cdot 3^{x} - 27$

Let $\rm 3^{x} = a$, we get,

$\rm or, (a)^{2} = 12 \cdot a - 27$

The above equation is a quadratic in a. We use the mid-term factorization method to find its roots, we get,

$\rm or, a^{2} - 12 a + 27 = 0$

$\rm or, a^{2} - (3 + 9)a + 27 = 0$

$\rm or, a^{2} - 3a - 9a + 27 = 0$

$\rm or, a(a - 3) - 9(a - 3) = 0$

$\rm or, (a - 9)(a - 3) = 0$

Either

$\rm (a - 9) = 0$

$\rm or, a = 9$

By our supposition earlier, $\rm a = 3^{x}$, we get,

$\rm or, 3^{x} = 3^{2}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x =2$

Or

$\rm (a - 3) = 0$

$\rm or, a = 3$

By our supposition earlier, $\rm a = 3^{x}$, we get,

$\rm or, 3^{x} = 3^{1}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = 1$

Hence, the required values of x are x = {1,2}.