Exponential Inequality

We are given the inequality $2 ^x (2 \cdot 2 ^x + 8) \leq 8 ^x (5 - 2 ^x)$ . Substituting $y = 2^x$ , we are asked to find all $x$ which satisfy $W(y) \leq 0$ , where:

$\begin{aligned} W(y) &= y(2y + 8) - y ^3 (5 - y) \\ &= y(2y + 8 - y^2(5 - y)) \\ &= y(y^3 - 5y^2 + 2y + 8) \\ &= y(y + 1)(y - 2)(y - 4). \end{aligned}$

From this factorization, given $y = 2^x > 0$ , only four intervals exist where $W(y)$ differs in sign. Testing each yields:

• $y \in (-\infty, -1) \implies W(y) \leq 0$
• $y \in (-1, 2) \implies W(y) \geq 0$
• $y \in (2, 4) \implies W(y) \leq 0$
• $y \in (4, \infty) \implies W(y) \geq 0$

As desired, $W(y) \leq 0$ in two of the four intervals: $(-\infty, -1)$ and $(2, 4)$ . However, since $y = 2^x > 0$ , $y$ must also be positive. Hence, $y \in (2, 4)$ .

So, the bounds are given by $2 = 2^x$ and $4 = 2^x$ , i.e. $x = 1$ and $x = 2$ . The sum of these bounds is $1 + 2 = 3$ .