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

We need to solve the inequality $2^x + 2^{|x|} \ge 2\sqrt{2}$. We consider two cases for $|x|$:

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

Case 2: $x < 0$ If $x < 0$, then $|x| = -x$. 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: $y + \frac{1}{y} \ge 2\sqrt{2}$ Multiply by $y$ (which is positive, so the inequality direction is preserved): $y^2 + 1 \ge 2\sqrt{2}y$ Rearrange into a quadratic inequality: $y^2 - 2\sqrt{2}y + 1 \ge 0$ To find the roots of $y^2 - 2\sqrt{2}y + 1 = 0$, we use the quadratic formula: $y = \frac{2\sqrt{2} \pm \sqrt{(-2\sqrt{2})^2 - 4(1)(1)}}{2} = \frac{2\sqrt{2} \pm \sqrt{8 - 4}}{2} = \frac{2\sqrt{2} \pm \sqrt{4}}{2} = \frac{2\sqrt{2} \pm 2}{2} = \sqrt{2} \pm 1$. The roots are $y_1 = \sqrt{2} - 1$ and $y_2 = \sqrt{2} + 1$. Since the quadratic has a positive leading coefficient, $y^2 - 2\sqrt{2}y + 1 \ge 0$ when $y \le \sqrt{2} - 1$ or $y \ge \sqrt{2} + 1$.

Now we apply the condition $0 < y < 1$:

• $y \le \sqrt{2} - 1$: Since $\sqrt{2} - 1 \approx 0.414$, this condition is compatible with $0 < y < 1$. 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 is consistent with $x < 0$ as $\log_2(\sqrt{2} - 1) < 0$. The solution from this part is $(-\infty, \log_2(\sqrt{2} - 1)]$.
• $y \ge \sqrt{2} + 1$: Since $\sqrt{2} + 1 \approx 2.414$, this condition ($y \ge 2.414$) contradicts $0 < y < 1$. Thus, there is no solution from this part for $x < 0$.

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