Basic Mathematics & Logarithm
Properties of Logarithms
Grade 11
Question:
<p>Which is the correct order for a given number <i>a</i>, <i>a</i> > 1:</p>
<p>(a) \(\log_2 a < \log_3 a < \log_e a < \log_{10} a\)</p>
<p>(b) \(\log_{10} a < \log_3 a < \log_e a < \log_2 a\)</p>
<p>(c) \(\log_{10} a < \log_e a < \log_2 a < \log_3 a\)</p>
<p>(d) \(\log_3 a < \log_e a < \log_2 a < \log_{10} a\)</p>
Step-by-Step Solution
Key Concept: When the base of a logarithm increases, the logarithmic value decreases for the same argument a > 1. This is because log_b(a) = ln(a)/ln(b), and as the base b increases, ln(b) increases, making the ratio smaller.
<p><strong>Step 1:</strong> Recall the change of base formula: log_b(a) = ln(a)/ln(b)</p><p><strong>Step 2:</strong> For a > 1, we have ln(a) > 0 (constant numerator). As the base b increases, ln(b) increases, so the fraction ln(a)/ln(b) decreases.</p><p><strong>Step 3:</strong> Order the bases: 2 < 3 < 10. Therefore: ln(2) < ln(3) < ln(10)</p><p><strong>Step 4:</strong> Since the numerator ln(a) is fixed and positive, we have: log_2(a) > log_3(a) > log_10(a)</p><p><strong>Step 5:</strong> The correct ordering from smallest to largest is: log_10(a) < log_3(a) < log_2(a)</p><p><strong>Step 6:</strong> This matches option (c): log_10(a) < log_3(a) < log_2(a)</p><p><strong>∴ Answer:</strong> c</p>
Correct Answer: c