Compute the sum of all real numbers x which satisfy the following equation $$\frac {8^x - 19 \cdot 4^x}{16 - 25 \cdot 2^x}= 2$$
1. **Substitution**: Let \( a = 2^x \). Then the given equation transforms as follows:
   \[
   \frac{8^x - 19 \cdot 4^x}{16 - 25 \cdot 2^x} = 2
   \]
   Since \( 8^x = (2^3)^x = (2^x)^3 = a^3 \) and \( 4^x = (2^2)^x = (2^x)^2 = a^2 \), the equation becomes:
   \[
   \frac{a^3 - 19a^2}{16 - 25a} = 2
   \]

2. **Simplification**: Multiply both sides of the equation by the denominator to clear the fraction:
   \[
   a^3 - 19a^2 = 2(16 - 25a)
   \]
   Simplify the right-hand side:
   \[
   a^3 - 19a^2 = 32 - 50a
   \]

3. **Rearrange the equation**: Move all terms to one side to set the equation to zero:
   \[
   a^3 - 19a^2 + 50a - 32 = 0
   \]

4. **Factor the polynomial**: We need to factor the cubic polynomial \( a^3 - 19a^2 + 50a - 32 \). By inspection or using the Rational Root Theorem, we find that \( a = 1 \), \( a = 2 \), and \( a = 16 \) are roots. Thus, the polynomial can be factored as:
   \[
   (a - 1)(a - 2)(a - 16) = 0
   \]

5. **Solve for \( a \)**: Set each factor equal to zero to find the values of \( a \):
   \[
   a - 1 = 0 \implies a = 1
   \]
   \[
   a - 2 = 0 \implies a = 2
   \]
   \[
   a - 16 = 0 \implies a = 16
   \]

6. **Back-substitute \( a = 2^x \)**: Solve for \( x \) in each case:
   \[
   2^x = 1 \implies x = 0
   \]
   \[
   2^x = 2 \implies x = 1
   \]
   \[
   2^x = 16 \implies x = 4
   \]

7. **Sum of all solutions**: Add the solutions \( x = 0 \), \( x = 1 \), and \( x = 4 \):
   \[
   0 + 1 + 4 = 5
   \]

The final answer is \(\boxed{5}\).