If `log_(10)2,log_(10)(2^(x)-1)` and `log_(10)(2^(x)+3)` are in A.P then the value of x is

Options

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

Answer

To solve the problem where \( \log_{10} 2, \log_{10} (2^x - 1), \) and \( \log_{10} (2^x + 3) \) are in Arithmetic Progression (A.P.), we can follow these steps:

### Step 1: Set up the A.P. condition
For three numbers \( a, b, c \) to be in A.P., the condition is:
\[ 2b = a + c \]
Here, let:
- \( a = \log_{10} 2 \)
- \( b = \log_{10} (2^x - 1) \)
- \( c = \log_{10} (2^x + 3) \)

Substituting these into the A.P. condition gives:
\[ 2 \log_{10} (2^x - 1) = \log_{10} 2 + \log_{10} (2^x + 3) \]

### Step 2: Use properties of logarithms
Using the property of logarithms that states \( \log_a b + \log_a c = \log_a (bc) \), we can rewrite the right-hand side:
\[ 2 \log_{10} (2^x - 1) = \log_{10} (2 \cdot (2^x + 3)) \]

### Step 3: Eliminate the logarithm
Since the logarithm function is one-to-one, we can exponentiate both sides to eliminate the logarithm:
\[ (2^x - 1)^2 = 2(2^x + 3) \]

### Step 4: Expand and simplify
Expanding both sides gives:
\[ (2^x - 1)(2^x - 1) = 2(2^x + 3) \]
\[ (2^x)^2 - 2 \cdot 2^x + 1 = 2 \cdot 2^x + 6 \]
\[ (2^x)^2 - 2 \cdot 2^x - 2 \cdot 2^x - 6 + 1 = 0 \]
\[ (2^x)^2 - 4 \cdot 2^x - 5 = 0 \]

### Step 5: Substitute \( t = 2^x \)
Let \( t = 2^x \). The equation becomes:
\[ t^2 - 4t - 5 = 0 \]

### Step 6: Factor the quadratic equation
Factoring gives:
\[ (t - 5)(t + 1) = 0 \]
Thus, we have two possible solutions:
\[ t - 5 = 0 \quad \text{or} \quad t + 1 = 0 \]
This leads to:
\[ t = 5 \quad \text{or} \quad t = -1 \]

### Step 7: Solve for \( x \)
Since \( t = 2^x \) cannot be negative, we discard \( t = -1 \). Therefore, we have:
\[ 2^x = 5 \]
Taking logarithm base 2 on both sides gives:
\[ x = \log_2 5 \]

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{\log_2 5} \]