If `log_(10)2,log_(10)(2^(x)-1) and log_(10)(2^(x)+3)` are three consecutive terms of an AP, then the value of x is

Options

A. 1
B. `log_(5)2`
C. `log_(2)5`
D. `log_(10)5`

Answer

To solve the problem, we need to determine the value of \( x \) such that \( \log_{10} 2 \), \( \log_{10} (2^x - 1) \), and \( \log_{10} (2^x + 3) \) are three consecutive terms of an arithmetic progression (AP).

### Step 1: Set Up the Condition for AP
For three numbers \( a \), \( b \), and \( c \) to be in AP, the condition is:
\[ 2b = a + c \]
In our case:
- \( a = \log_{10} 2 \)
- \( b = \log_{10} (2^x - 1) \)
- \( c = \log_{10} (2^x + 3) \)

Thus, we can write:
\[ 2 \log_{10} (2^x - 1) = \log_{10} 2 + \log_{10} (2^x + 3) \]

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

### Step 3: Rewrite the Equation
Using the property \( n \log_a b = \log_a (b^n) \), we can rewrite the left-hand side:
\[ \log_{10} ((2^x - 1)^2) = \log_{10} (2(2^x + 3)) \]

### Step 4: Remove the Logarithm
Since the logarithms are equal, we can set the arguments equal to each other:
\[ (2^x - 1)^2 = 2(2^x + 3) \]

### Step 5: Expand Both Sides
Expanding both sides gives:
\[ (2^x - 1)(2^x - 1) = 2 \cdot 2^x + 6 \]
\[ (2^x)^2 - 2 \cdot 2^x + 1 = 2 \cdot 2^x + 6 \]

### Step 6: Rearrange the Equation
Rearranging the equation leads to:
\[ (2^x)^2 - 4 \cdot 2^x - 5 = 0 \]

### Step 7: Factor the Quadratic
This quadratic can be factored as:
\[ (2^x - 5)(2^x + 1) = 0 \]

### Step 8: Solve for \( 2^x \)
Setting each factor to zero gives:
1. \( 2^x - 5 = 0 \) → \( 2^x = 5 \) → \( x = \log_2 5 \)
2. \( 2^x + 1 = 0 \) → (This has no real solution since \( 2^x \) is always positive)

### Conclusion
Thus, the only valid solution is:
\[ x = \log_2 5 \]