If `4^(x + 1) - 16^(x) gt 2log_(4)8` then `x` may be

Options

A. `(1)/(2)`
B. `(3)/(2)`
C. `(3)/(4)`
D. `(4)/(3)`

Explanation

To solve the inequality \( 4^{(x + 1)} - 16^{x} > 2\log_{4}8 \), we will follow these steps:

### Step 1: Rewrite the terms in the inequality
We know that \( 16 = 4^2 \), so we can rewrite \( 16^x \) as \( (4^2)^x = 4^{2x} \). Thus, the inequality becomes:
\[ 4^{(x + 1)} - 4^{(2x)} > 2\log_{4}8 \]

### Step 2: Simplify the left-hand side
Now, we can express \( 4^{(x + 1)} \) as \( 4^x \cdot 4^1 = 4 \cdot 4^x \). Therefore, the inequality can be rewritten as:
\[ 4 \cdot 4^x - 4^{(2x)} > 2\log_{4}8 \]

### Step 3: Simplify the right-hand side
Next, we simplify \( 2\log_{4}8 \). We know that \( 8 = 2^3 \) and \( 4 = 2^2 \), so:
\[ \log_{4}8 = \frac{\log_{2}8}{\log_{2}4} = \frac{3}{2} \quad \text{(since } \log_{2}8 = 3 \text{ and } \log_{2}4 = 2\text{)} \]
Thus,
\[ 2\log_{4}8 = 2 \cdot \frac{3}{2} = 3 \]

### Step 4: Substitute back into the inequality
Now substituting back, we have:
\[ 4 \cdot 4^x - 4^{(2x)} > 3 \]

### Step 5: Rearranging the inequality
Rearranging gives us:
\[ 4 \cdot 4^x - 3 > 4^{(2x)} \]

### Step 6: Let \( y = 4^x \)
Let \( y = 4^x \). Then the inequality becomes:
\[ 4y - 3 > y^2 \]
or rearranging it:
\[ y^2 - 4y + 3 < 0 \]

### Step 7: Factor the quadratic
Factoring the quadratic gives us:
\[ (y - 1)(y - 3) < 0 \]

### Step 8: Determine the intervals
To find the intervals where this inequality holds, we can analyze the sign of the product:
- The roots are \( y = 1 \) and \( y = 3 \).
- The inequality \( (y - 1)(y - 3) < 0 \) holds between the roots:
\[ 1 < y < 3 \]

### Step 9: Substitute back for \( y \)
Since \( y = 4^x \), we have:
\[ 1 < 4^x < 3 \]

### Step 10: Taking logarithms
Taking logarithm base 4, we get:
\[ \log_{4}1 < x < \log_{4}3 \]
Since \( \log_{4}1 = 0 \), we have:
\[ 0 < x < \log_{4}3 \]

### Final Answer
Thus, the solution for \( x \) is:
\[ x \in \left(0, \log_{4}3\right) \]