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

To determine the correct order of the logarithmic expressions for a given positive number \(\alpha\), we need to understand the behavior of logarithmic functions with different bases. Here, the options involve logarithms with bases 2, 3, \(e\), and 10. 

The logarithmic function \(\log_b(\alpha)\) is a monotonically increasing function for \(\alpha > 0\) and \(b > 1\). Thus, if you have a fixed positive number \(\alpha\), the behavior of the logarithm is entirely dependent on the base \(b\):

• If the base \(b\) is smaller, \(\log_b(\alpha)\) is larger, because a smaller base means the logarithm grows faster to reach the value of \(\alpha\).
• If the base \(b\) is larger, \(\log_b(\alpha)\) is smaller, as the growth of the logarithm is slower.

For the bases involved:

• \(\log_2\) has the smallest base, so it will have the largest value for the same \(\alpha\).
• \(\log_3\) is next, so it will be smaller than \(\log_2\).
• \(\log_e\), also known as the natural logarithm, falls between \(\log_3\) and \(\log_{10}\).
• \(\log_{10}\) has the largest base of all, so it will have the smallest value for \(\alpha\).

Therefore, the correct ascending order of the logarithms for a fixed positive number \(\alpha\) is:

• \(\log_{10} \alpha\) (smallest)
• \(\log_3 \alpha\)
• \(\log_e \alpha\)
• \(\log_2 \alpha\) (largest)

Based on the options provided, the correct order is: \(\log_{10} \alpha, \log_3 \alpha, \log_e \alpha, \log_2 \alpha\).