Given

$\rm 2^{x+3} + \frac{1}{2^{x}} - 9 = 0$

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

$\rm or, 2^{x} \cdot 2^{3} + \frac{1}{2^{x}} - 9 = 0$

$\rm or, 2^{x} \cdot 8 + \frac{1}{2^{x}} - 9 = 0$

Multiplying both sides of the equation by $\rm 2^{x}$, we get,

$\rm or, 2^{x} \cdot 2^{x} \cdot 8 + 2^{x} \cdot \frac{1}{2^{x}} - 2^{x} \cdot 9 = 0$

$\rm or, 2^{x} \cdot 2^{x} \cdot 8 + 1 - 2^{x} \cdot 9 = 0$

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

$\rm or, 2^{x + x} \cdot 8 + 1 - 2^{x} \cdot 9 = 0$

$\rm or, 2^{2x} \cdot 8 + 1 - 2^{x} \cdot 9 = 0$

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

$\rm or, \left ( 2^{x} \right ) ^{2} \cdot 8 + 1 - 2^{x} \cdot 9 = 0$

Let $\rm 2^{x} = a$. Substituting the supposed values in the above equation, we get,

$\rm or, a^{2} \cdot 8 + 1 - a \cdot 9$

The above equation is quadratic in a. We use the mid-term factorization method to solve the quadratic equation.

$\rm or, 8 a^{2} - 9 a + 1 = 0$

$\rm or, 8 a^{2} - ( 8 + 1) a + 1 = 0$

$\rm or, 8a^{2} - 8a - a + 1 = 0$

$\rm or, 8a ( a - 1) - 1( a- 1) = 0$

$\rm or, (8a - 1)(a - 1) = 0$

Either

$\rm (8a - 1) = 0$

$\rm or, 8a = 1$

$\rm or, a = \frac{1}{8}$

$\rm or, a = \frac{1}{2^{3}}$

By the law of indices, $\rm \frac{1}{a^{m}} = a^{-m}$,

$\rm or, a = 2^{-3}$

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

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

$\rm \therefore x = -3$

Or

$\rm (a - 1) = 0$

$\rm or, a = 1$

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

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

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

$\rm \therefore x = 0$

Hence, the required values of x are x = {-3, 0}.