The set of all real numbers satisfying the inequation
`2^(x)+2^(|x|) gt 2sqrt(2)`, is

Options

A. `(1//2,oo)`
B. `(-oo,log_(2)(sqrt(2)-1)`
C. `(-oo,1//2)`
D. `[1//2,oo) cup (-oo,log_(2)(sqrt(2)-1))`

Answer

To solve the inequality \( 2^x + 2^{|x|} > 2\sqrt{2} \), we will consider two cases based on the value of \( x \): when \( x \geq 0 \) and when \( x < 0 \).

### Step 1: Case 1 - \( x \geq 0 \)

In this case, \( |x| = x \). Therefore, the inequality becomes:
\[ 2^x + 2^x > 2\sqrt{2} \]
This simplifies to:
\[ 2 \cdot 2^x > 2\sqrt{2} \]
Dividing both sides by 2:
\[ 2^x > \sqrt{2} \]
We can rewrite \( \sqrt{2} \) as \( 2^{1/2} \):
\[ 2^x > 2^{1/2} \]
Since the bases are the same, we can compare the exponents:
\[ x > \frac{1}{2} \]

### Step 2: Case 2 - \( x < 0 \)

In this case, \( |x| = -x \). Therefore, the inequality becomes:
\[ 2^x + 2^{-x} > 2\sqrt{2} \]
Let \( k = 2^x \). Since \( x < 0 \), \( k \) will be positive and less than 1. The inequality can be rewritten as:
\[ k + \frac{1}{k} > 2\sqrt{2} \]
Multiplying through by \( k \) (since \( k > 0 \)):
\[ k^2 + 1 > 2\sqrt{2}k \]
Rearranging gives:
\[ k^2 - 2\sqrt{2}k + 1 > 0 \]

### Step 3: Solving the Quadratic Inequality

The quadratic \( k^2 - 2\sqrt{2}k + 1 = 0 \) can be solved using the quadratic formula:
\[ k = \frac{2\sqrt{2} \pm \sqrt{(2\sqrt{2})^2 - 4 \cdot 1}}{2} \]
Calculating the discriminant:
\[ (2\sqrt{2})^2 - 4 = 8 - 4 = 4 \]
Thus, the roots are:
\[ k = \frac{2\sqrt{2} \pm 2}{2} = \sqrt{2} \pm 1 \]
This gives us two roots:
\[ k_1 = \sqrt{2} - 1 \quad \text{and} \quad k_2 = \sqrt{2} + 1 \]

### Step 4: Analyzing the Quadratic

The quadratic \( k^2 - 2\sqrt{2}k + 1 \) opens upwards (since the coefficient of \( k^2 \) is positive). Therefore, it is positive outside the interval defined by its roots:
\[ k < \sqrt{2} - 1 \quad \text{or} \quad k > \sqrt{2} + 1 \]
Since \( k = 2^x \) and \( k < 1 \) (because \( x < 0 \)), we only consider:
\[ k < \sqrt{2} - 1 \]

### Step 5: Finding the Corresponding \( x \)

Now we need to find \( x \) such that:
\[ 2^x < \sqrt{2} - 1 \]
Taking logarithm base 2:
\[ x < \log_2(\sqrt{2} - 1) \]

### Step 6: Final Solution

Combining both cases, we have:
1. From Case 1: \( x > \frac{1}{2} \)
2. From Case 2: \( x < \log_2(\sqrt{2} - 1) \)

Thus, the solution set is:
\[ (-\infty, \log_2(\sqrt{2} - 1)) \cup \left(\frac{1}{2}, \infty\right) \]