If `x in R`, the solution ste of the equation `4^(-x+0.5)-7.2^(-x)-4lt0` is equal to

Options

A. `(2,7/2)`
B. `(-2,oo)`
C. `(2,oo)`
D. `(-oo,oo)`

Answer

To solve the inequality \( 4^{-x + 0.5} - 7 \cdot 2^{-x} - 4 < 0 \), we will follow these steps:

### Step 1: Rewrite the terms with base 2
We know that \( 4 = 2^2 \). Therefore, we can rewrite \( 4^{-x + 0.5} \) as:
\[ 4^{-x + 0.5} = (2^2)^{-x + 0.5} = 2^{-2x + 1} = 2^{1 - 2x} \]
Now, substituting this back into the inequality gives us:
\[ 2^{1 - 2x} - 7 \cdot 2^{-x} - 4 < 0 \]

### Step 2: Substitute \( y = 2^{-x} \)
Let \( y = 2^{-x} \). Then \( 2^{1 - 2x} = 2 \cdot y^2 \). The inequality now becomes:
\[ 2y^2 - 7y - 4 < 0 \]

### Step 3: Factor the quadratic
We need to factor the quadratic \( 2y^2 - 7y - 4 \). To do this, we can use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2, b = -7, c = -4 \):
\[ y = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot (-4)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ 49 + 32 = 81 \]
So,
\[ y = \frac{7 \pm 9}{4} \]
This gives us two solutions:
\[ y_1 = \frac{16}{4} = 4 \quad \text{and} \quad y_2 = \frac{-2}{4} = -\frac{1}{2} \]

### Step 4: Analyze the intervals
The roots of the quadratic are \( y = 4 \) and \( y = -\frac{1}{2} \). Since \( y = 2^{-x} \) must be positive, we only consider \( y = 4 \). The quadratic \( 2y^2 - 7y - 4 \) opens upwards (since the coefficient of \( y^2 \) is positive), so it is negative between the roots:
\[ -\frac{1}{2} < y < 4 \]
Since \( y \) must be positive, we have:
\[ 0 < y < 4 \]

### Step 5: Substitute back for \( x \)
Recall that \( y = 2^{-x} \). Therefore, we have:
\[ 0 < 2^{-x} < 4 \]
Taking logarithm base 2:
\[ 0 < -x < 2 \quad \Rightarrow \quad -2 < x < 0 \]

### Final Solution
The solution set for the inequality \( 4^{-x + 0.5} - 7 \cdot 2^{-x} - 4 < 0 \) is:
\[ (-2, 0) \]