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Problem 6 (3 marks). Without the use of calculators or software establish which number is larger: log_(189)1323 or log_(63)147.

Without the use of calculators or software establish which number is larger: \newlinelog⁡1891323\log_{189}1323 or log⁡63147\log_{63}147.

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Full solution

Q. Without the use of calculators or software establish which number is larger: \newlinelog⁡1891323\log_{189}1323 or log⁡63147\log_{63}147.
  1. Use Change of Base Formula: Use the change of base formula to compare the two logarithms.\newlinelog⁡1891323=log⁡(1323)log⁡(189)\log_{189}1323 = \frac{\log(1323)}{\log(189)}\newlinelog⁡63147=log⁡(147)log⁡(63)\log_{63}147 = \frac{\log(147)}{\log(63)}
  2. Simplify Bases and Numbers: Simplify the bases and the numbers inside the logarithms by finding common factors.\newline189=33×7189 = 3^3 \times 7\newline63=32×763 = 3^2 \times 7\newline1323=33×721323 = 3^3 \times 7^2\newline147=3×72147 = 3 \times 7^2
  3. Cancel Common Factors: Simplify the expressions by canceling out common factors.\newlinelog⁡1891323=log⁡(33⋅72)log⁡(33⋅7)\log_{189} 1323 = \frac{\log(3^3 \cdot 7^2)}{\log(3^3 \cdot 7)}\newlinelog⁡63147=log⁡(3⋅72)log⁡(32⋅7)\log_{63} 147 = \frac{\log(3 \cdot 7^2)}{\log(3^2 \cdot 7)}
  4. Apply Power Rule: Use the power rule of logarithms to bring down the exponents.\newlinelog⁡1891323=3⋅log⁡(7)+2⋅log⁡(7)3⋅log⁡(3)+log⁡(7)\log_{189}1323 = \frac{3 \cdot \log(7) + 2 \cdot \log(7)}{3 \cdot \log(3) + \log(7)}\newlinelog⁡63147=log⁡(3)+2⋅log⁡(7)2⋅log⁡(3)+log⁡(7)\log_{63}147 = \frac{\log(3) + 2 \cdot \log(7)}{2 \cdot \log(3) + \log(7)}
  5. Combine Terms: Combine the terms with common logarithms.\newlinelog⁡1891323=5⋅log⁡(7)3⋅log⁡(3)+log⁡(7)\log_{189}1323 = \frac{5 \cdot \log(7)}{3 \cdot \log(3) + \log(7)}\newlinelog⁡63147=2⋅log⁡(7)+log⁡(3)2⋅log⁡(3)+log⁡(7)\log_{63}147 = \frac{2 \cdot \log(7) + \log(3)}{2 \cdot \log(3) + \log(7)}
  6. Compare Numerators and Denominators: Compare the numerators and denominators of both expressions.\newlineFor log⁡1891323\log_{189}1323, the numerator is 5×log⁡(7)5 \times \log(7) and the denominator is 3×log⁡(3)+log⁡(7)3 \times \log(3) + \log(7).\newlineFor log⁡63147\log_{63}147, the numerator is 2×log⁡(7)+log⁡(3)2 \times \log(7) + \log(3) and the denominator is 2×log⁡(3)+log⁡(7)2 \times \log(3) + \log(7).
  7. Correct Comparison Mistake: Notice that the numerator of log⁡1891323\log_{189}1323 is larger than the numerator of log⁡63147\log_{63}147, since 5 \times \log(7) > 2 \times \log(7) + \log(3). However, the denominator of log⁡1891323\log_{189}1323 is also larger than the denominator of log⁡63147\log_{63}147, since 3 \times \log(3) + \log(7) > 2 \times \log(3) + \log(7).
  8. Correct Comparison Mistake: Notice that the numerator of log⁡1891323\log_{189}1323 is larger than the numerator of log⁡63147\log_{63}147, since 5 \times \log(7) > 2 \times \log(7) + \log(3). However, the denominator of log⁡1891323\log_{189}1323 is also larger than the denominator of log⁡63147\log_{63}147, since 3 \times \log(3) + \log(7) > 2 \times \log(3) + \log(7). Since both the numerator and denominator of log⁡1891323\log_{189}1323 are larger than those of log⁡63147\log_{63}147, we cannot directly conclude which fraction is larger without further simplification or numerical evaluation.
  9. Correct Comparison Mistake: Notice that the numerator of log⁡1891323\log_{189}1323 is larger than the numerator of log⁡63147\log_{63}147, since 5 \times \log(7) > 2 \times \log(7) + \log(3). However, the denominator of log⁡1891323\log_{189}1323 is also larger than the denominator of log⁡63147\log_{63}147, since 3 \times \log(3) + \log(7) > 2 \times \log(3) + \log(7). Since both the numerator and denominator of log⁡1891323\log_{189}1323 are larger than those of log⁡63147\log_{63}147, we cannot directly conclude which fraction is larger without further simplification or numerical evaluation. Realize that a mistake was made in the previous step. The comparison of the numerators and denominators was incorrect. We need to correct this and compare the values properly.

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