[yellow]

6)

$$2 + \sqrt{-(2^{2x} - 2^y)^2} = log_3 (x+y)$$

$$\therefore (2^{2x} - 2^y) = 0$$

$$ 2x = y$$



$$ 2 = log_3 (3x)$$

$$3^2 = 3x$$

$$x = 3$$

$$y = 6$$

$$x^2 + y = 3^2 + 6 = 15$$