If `log 2, log (2^(n) - 1) and log(2^(n) + 3)` are in A.P., then n =

Options

A. `log_(3)5`
B. `log_(2)5`
C. `(5)/(2)`
D. `(3)/(2)`

Answer

To solve the problem, we need to determine the value of \( n \) such that \( \log 2 \), \( \log(2^n - 1) \), and \( \log(2^n + 3) \) are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Understanding A.P. Condition**:
For three numbers \( a \), \( b \), and \( c \) to be in A.P., the condition is:
\[ 2b = a + c \]
In our case, let:
- \( a = \log 2 \)
- \( b = \log(2^n - 1) \)
- \( c = \log(2^n + 3) \)

Thus, we write:
\[ 2 \log(2^n - 1) = \log 2 + \log(2^n + 3) \]

2. **Using Logarithmic Properties**:
We can use the property of logarithms that states \( \log a + \log b = \log(ab) \). Therefore, we can rewrite the equation as:
\[ 2 \log(2^n - 1) = \log(2 \cdot (2^n + 3)) \]

3. **Exponentiating Both Sides**:
To eliminate the logarithm, we exponentiate both sides:
\[ (2^n - 1)^2 = 2(2^n + 3) \]

4. **Expanding Both Sides**:
Now we expand both sides:
\[ (2^n - 1)(2^n - 1) = 2 \cdot 2^n + 6 \]
This simplifies to:
\[ 2^{2n} - 2 \cdot 2^n + 1 = 2^{n+1} + 6 \]

5. **Rearranging the Equation**:
Rearranging gives us:
\[ 2^{2n} - 2^{n+1} - 2 \cdot 2^n + 1 - 6 = 0 \]
Simplifying further:
\[ 2^{2n} - 4 \cdot 2^n - 5 = 0 \]

6. **Substituting \( x = 2^n \)**:
Let \( x = 2^n \). The equation becomes:
\[ x^2 - 4x - 5 = 0 \]

7. **Factoring the Quadratic**:
We can factor the quadratic:
\[ (x - 5)(x + 1) = 0 \]
This gives us:
\[ x = 5 \quad \text{or} \quad x = -1 \]

8. **Finding \( n \)**:
Since \( x = 2^n \) and \( x \) must be positive, we take:
\[ 2^n = 5 \]
Taking logarithm base 2:
\[ n = \log_2 5 \]

### Final Answer:
Thus, the value of \( n \) is:
\[ n = \log_2 5 \]