Question

If \( \log_3 2, \log_3 (2^x - 5), \log_3 (2^x - 7/2) \) are in arithmetic progression, then the value of x is equal to

• A. 5
• B. 4
• C. 2
• D. 3

Hint:

For three terms in AP, twice the middle term equals the sum of the other two. Turn the logarithms into one equation in 2^x, and remember the argument of a logarithm must be positive.

Answer 1

The Correct Option is D

Step 1: Condition for an AP.
Three terms \( p, q, r \) are in AP when \( 2q = p + r \). Here \( p = \log_3 2 \), \( q = \log_3 (2^x - 5) \) and \( r = \log_3 (2^x - \frac{7}{2}) \).

Step 2: Write the condition.
\[ 2\log_3 (2^x - 5) = \log_3 2 + \log_3 \left(2^x - \frac{7}{2}\right) \]
Use \( 2\log a = \log a^2 \) on the left and \( \log a + \log b = \log ab \) on the right:
\[ \log_3 (2^x - 5)^2 = \log_3 \left[2\left(2^x - \frac{7}{2}\right)\right] \]

Step 3: Remove the logarithms.
Both sides have the same base, so the arguments are equal:
\[ (2^x - 5)^2 = 2 \cdot 2^x - 7 \]
Let \( t = 2^x \). Then \( t^2 - 10t + 25 = 2t - 7 \).

Step 4: Solve the quadratic.
\[ t^2 - 12t + 32 = 0 \quad \Rightarrow \quad (t - 4)(t - 8) = 0 \]
So \( t = 4 \) or \( t = 8 \).

Step 5: Apply the domain condition.
A logarithm needs a positive argument, so \( 2^x - 5 > 0 \), that is \( t > 5 \). The value \( t = 4 \) gives \( 4 - 5 = -1 \), which is not allowed. So \( t = 8 \), meaning \( 2^x = 8 \) and \( x = 3 \).

Step 6: Check and reject other options.
For \( x = 3 \) the terms are \( \log_3 2 \), \( \log_3 3 = 1 \) and \( \log_3 4.5 \). Then \( 2 \times 1 = 2 \) and \( \log_3 2 + \log_3 4.5 = \log_3 9 = 2 \). Equal, so it works.
For \( x = 2 \), the term \( 2^2 - 5 = -1 \) is negative, so the logarithm is undefined. For \( x = 4 \) and \( x = 5 \), the squares \( 11^2 = 121 \) and \( 27^2 = 729 \) do not match \( 2 \times 12.5 = 25 \) and \( 2 \times 28.5 = 57 \).

Final Answer:
The value of x is 3, which is option 4. \[ \boxed{4} \]