Solue \( 70 r x \) \( 2^{x}+3 \cdot 2^{x+1}=56 \) \( 4^{x}-9<5 x \)

To start solving the equation \( 2^{x} + 3 \cdot 2^{x+1} = 56 \), we can rewrite it as \( 2^{x} + 3 \cdot 2 \cdot 2^{x} = 56 \), which simplifies to \( 2^{x} (1 + 6) = 56 \) or \( 7 \cdot 2^{x} = 56 \). Dividing both sides by 7 gives us \( 2^{x} = 8 \), leading to \( x = 3 \). Next, let's tackle the inequality \( 4^{x} - 9 < 5x \). We can express \( 4^{x} \) as \( (2^{2})^{x} = (2^{x})^{2} \). This transforms our inequality into \( (2^{x})^{2} < 5x + 9 \). Approximating solutions by testing values of \( x \), we check \( x = 2 \) (which gives us \( 16 - 9 < 10 \), or \( 7 < 10 \) - true), and keep testing until \( x = 4 \) (where \( 16 - 9 < 20 \), or \( 7 < 20 \), which is still true). As we go higher, we find that beyond \( x = 5 \), the left-hand side grows faster than \( 5x + 9 \). Thus the inequality holds for \( x < 4 \). Keep checking for values until you find the largest valid \( x \). There you have it! Happy solving!