Question

\(\log_{10} 2\), \(\log_{10}\left(2^{x}-1\right)\) and \(\log_{10}\left(2^{x}+3\right)\) are three consecutive terms of an A. P. for :

• A. no real \(x\)
• B. exactly one real \(x\)
• C. exactly two real \(x\)
• D. more than two real \(x\).

Hint:

Three terms a, b, c are in A.P. when 2b = a + c; use this with logarithm rules and remember the log arguments must be positive.

Answer 1

The Correct Option is B

For three terms in A.P., the middle term is the average of the other two, so \(2\log_{10}(2^x-1)=\log_{10}2+\log_{10}(2^x+3)\).
Hence \((2^x-1)^2=2(2^x+3)\).
Let \(t=2^x>0\). Then \(t^2-2t+1=2t+6\), i.e. \(t^2-4t-5=0\), so \((t-5)(t+1)=0\).
\(t=-1\) is impossible because \(2^x>0\). Also \(2^x-1>0\) is needed for the logarithm, and \(t=5\) satisfies it.
So \(2^x=5\), giving \(x=\log_2 5\), exactly one real value.
Answer: (2) exactly one real \(x\)