Question

Which is the correct order for a given number \(\alpha\) in increasing order?

• A. \(\log_2 \alpha, \log_3 \alpha, \log_e \alpha, \log_{10} \alpha\)
• B. \(\log_{10} \alpha, \log_3 \alpha, \log_e \alpha, \log_2 \alpha\)
• C. \(\log_{10} \alpha, \log_e \alpha, \log_2 \alpha, \log_3 \alpha\)
• D. None of the above

Hint:

For \(\alpha>1\), larger base → smaller logarithm. Bases: \(2<e<3<10\) \(\rightarrow\) logs: \(\log_{10}<\log_3<\log_e<\log_2\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For \(\alpha>1\), \(\log_a \alpha\) is a decreasing function of the base \(a\).
Step 2: Detailed Explanation:
The bases in increasing order are: \(2<e<3<10\)
Since the base increases, the logarithm decreases.
Therefore, \(\log_{10} \alpha<\log_3 \alpha<\log_e \alpha<\log_2 \alpha\)
Thus the increasing order is: \(\log_{10} \alpha, \log_3 \alpha, \log_e \alpha, \log_2 \alpha\)
Step 3: Final Answer:
\(\log_{10} \alpha, \log_3 \alpha, \log_e \alpha, \log_2 \alpha\).