If `"log"_(2) 7 = x`, then x is:

Options

A. a rational number such that `0 lt x lt 2`
B. an irrational number such that `2 lt x lt3`
C. a rational number such that `2 lt x lt 3`
D. a prime number of the form `7x+2`

Answer

To solve the equation \( \log_2 7 = x \), we will follow these steps:

### Step 1: Rewrite the logarithmic expression
Using the property of logarithms, we can rewrite the equation:
\[ \log_b a = c \implies a = b^c \]
Applying this property to our expression, we have:
\[ 7 = 2^x \]

### Step 2: Identify the range of \( x \)
Next, we need to determine the range of \( x \) by comparing \( 7 \) with powers of \( 2 \):
- \( 2^2 = 4 \)
- \( 2^3 = 8 \)

Since \( 7 \) is between \( 4 \) and \( 8 \), we can conclude:
\[ 2 < x < 3 \]

### Step 3: Determine if \( x \) is rational or irrational
The value of \( x \) lies between \( 2 \) and \( 3 \). However, since \( 7 \) is not a power of \( 2 \), the exact value of \( x \) cannot be expressed as a simple fraction. Therefore, \( x \) is an irrational number.

### Final Conclusion
Thus, we conclude that:
\[ x \text{ is an irrational number such that } 2 < x < 3. \]