Solve the inequality 1 + 11/(2 to the x, minus 8) + 28/(4 to the x, minus 2 to the (x+4), plus 64) ≥ 0.

Step-by-step solution

1. 1

   Substitute t = 2ˣ and rewrite every exponential in terms of t. Since $2^x > 0$ for all real $x$, set $t = 2^x$ with $t > 0$. Then

   $4^x = (2^x)^2 = t^2, \qquad 2^{x+4} = 2^4 \cdot 2^x = 16t$

   The key rewrite is $2^{x+4} = 16t$, not $t + 16$ — the exponent adds, so the base multiplies.

2. 2

   Recognise the second denominator as a perfect square.

   $4^x - 2^{x+4} + 64 = t^2 - 16t + 64 = (t - 8)^2$

   Both denominators now involve the same quantity $t - 8$, which is what makes the whole problem collapse. Write $u = t - 8$, so the inequality becomes

   $1 + \frac{11}{u} + \frac{28}{u^2} \ge 0, \qquad u \ne 0$

3. 3

   Multiply through by u² — legitimate because it is positive. Since $u \ne 0$, $u^2 > 0$, so multiplying does not flip the inequality:

   $u^2 + 11u + 28 \ge 0 \ \Longleftrightarrow \ (u + 4)(u + 7) \ge 0$

   This is why writing the denominators as a square and its root pays off: no sign chart on the fractions is needed.

4. 4

   Solve the quadratic inequality in u. An upward parabola is non-negative outside its roots $u = -7$ and $u = -4$:

   $u \le -7 \quad \text{or} \quad u \ge -4, \qquad u \ne 0$

5. 5

   Translate back through u = t − 8 and t = 2ˣ. From $u \le -7$:

   $t \le 1 \ \Rightarrow \ 2^x \le 2^0 \ \Rightarrow \ x \le 0$

   From $u \ge -4$:

   $t \ge 4 \ \Rightarrow \ 2^x \ge 2^2 \ \Rightarrow \ x \ge 2$

   Both use the fact that $2^x$ is increasing, so the direction of the inequality is preserved. The condition $t > 0$ is automatic and imposes nothing extra.

6. 6

   Remove the excluded point and state the answer. The restriction $u \ne 0$ means $t \ne 8$, i.e. $x \ne 3$, where both original denominators vanish. So

   $x \in (-\infty, 0] \cup [2, 3) \cup (3, \infty)$

   Numerical spot checks confirm it: at $x = -1$ the left side is $+0.0311$, at $x = 0$ it is exactly $0$, at $x = 1$ it is $-0.0556$, at $x = 2$ it is exactly $0$, and at $x = 4$ it is $+2.8125$.