Question

The solution of $2^x + 2^{|x|} \ge 2\sqrt{2}$ is

• A. $(-\infty, log_2(\sqrt{2}+1))$
• B. $(0, \infty)$
• C. $( \frac{1}{2}, log_2(\sqrt{2}-1))$
• D. $(-\infty, log_2(\sqrt{2}-1)] \cup [\frac{1}{2}, \infty)$

Answer 1

Answer: (D)

$(-\infty, log_2(\sqrt{2}-1)] \cup [\frac{1}{2}, \infty)$

Answer 2

The inequality $2^x + 2^{|x|} \ge 2\sqrt{2}$ is solved by considering two cases:

Case 1: $x \ge 0$ In this case, $|x| = x$. The inequality becomes: $2^x + 2^x \ge 2\sqrt{2}$ $2 \cdot 2^x \ge 2\sqrt{2}$ $2^{x+1} \ge 2^{3/2}$ Since the base $2 > 1$, we can compare the exponents: $x+1 \ge \frac{3}{2}$ $x \ge \frac{1}{2}$ The solution for this case is $[\frac{1}{2}, \infty)$.

Case 2: $x < 0$ In this case, $|x| = -x$. The inequality becomes: $2^x + 2^{-x} \ge 2\sqrt{2}$ Let $y = 2^x$. Since $x < 0$, we have $0 < y < 1$. The inequality in terms of $y$ is: $y + \frac{1}{y} \ge 2\sqrt{2}$ Multiplying by $y$ (which is positive): $y^2 + 1 \ge 2\sqrt{2}y$ $y^2 - 2\sqrt{2}y + 1 \ge 0$ The roots of the quadratic equation $y^2 - 2\sqrt{2}y + 1 = 0$ are $y = \sqrt{2} \pm 1$. Since the quadratic opens upwards, the inequality $y^2 - 2\sqrt{2}y + 1 \ge 0$ holds for $y \le \sqrt{2} - 1$ or $y \ge \sqrt{2} + 1$. Considering the constraint $0 < y < 1$:

• $y \le \sqrt{2} - 1$: This condition is satisfied since $\sqrt{2} - 1 \approx 0.414$, which is within $(0, 1)$. Substituting back $y = 2^x$: $2^x \le \sqrt{2} - 1$. Taking $\log_2$ on both sides: $x \le \log_2(\sqrt{2} - 1)$. This gives the interval $(-\infty, \log_2(\sqrt{2} - 1)]$.
• $y \ge \sqrt{2} + 1$: This condition is not satisfied for $y < 1$, as $\sqrt{2} + 1 \approx 2.414$.

Combining the Solutions The total solution is the union of the solutions from both cases: $(-\infty, \log_2(\sqrt{2} - 1)] \cup [\frac{1}{2}, \infty)$.