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How to compare \(\log_4 3,\log_5 4,\log_6 5.\).

Note by Kishan K
3 years, 9 months ago

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Lemma: If \(f(x) \) is a positive function that satisfies \( f(x+2) - f(x+1) < f(x+1) - f(x) \), then \( f(x+2)f(x) < f(x+1)^2 \)

Proof: Apply AM-GM.\(_\square\)

Note: This condition can be thought of as a concave function, in a discrete setting. This is all the calculus that is required.

Now apply this to \( f(x) = \ln x \) for \( x \geq 2 \). Calvin Lin Staff · 3 years, 9 months ago

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Show that \(\frac{\log(x)}{\log(x+1)}\) is an increasing function for \(x>0\), where the base is e. Abhishek Sinha · 3 years, 9 months ago

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@Abhishek Sinha I dont know much Calculus....please write ur solution.Is there any simple method......? Kishan K · 3 years, 9 months ago

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@Kishan K Let \( f(x) = \frac { \log (x) }{ \log (x + 1) } \) for \(x > 0 \)

By Quotient Rule, we have \( \large f'(x) = \frac { \log (x+1) \space \cdot \frac {1}{x} - \log (x) \space \cdot \frac {1}{x+1} } { ( \log (x+1))^2 } \) or

\( \large f'(x) = \frac { (x+1) \space \log (x+1) \space - x \space \log (x) } { x(x+1) \space ( \log (x+1))^2 } \)

\( \large f'(x) = \frac { x ( \log (x+1) - \log (x) ) + \log (x+1) } { x(x+1) ( \log (x+1))^2 } \)

Since for \( x > 0 \), we have \( \log (x+1) - \log (x) = \log ( 1 + \frac {1}{x} ) > 0 \) Thus the entire numerator of \( f'(x) \) is positive, same goes to the denominator of \(f'(x) \), thus \(f'(x) \) is strictly positive, or \(f(x) \) is an increasing function.

Thus \( f(3) < f(4) < f(5) \), we have \( \frac { \log 3 }{ \log 4 } < \frac { \log 4 }{ \log 5 } < \frac { \log 6 }{ \log 5 } \) or

\( \log_4 3 < \log_5 4 < \log_6 5 \)

Alternatively, suppose we consider their reciprocals \( \log_3 4, \log_4 5, \log_5 6 \)

\( \log_3 4 = \log_3 (3 * \frac {4}{3}), \log_4 5 = \log_4 (4 * \frac {5}{4}), \log_5 6 = \log_5 (5 * \frac {6}{5})\)

\( \log_3 4 = 1 + \log_3 \frac {4}{3}, \log_4 5 = 1 + \log_4 \frac {5}{4}, \log_5 6 = \log_5 \frac {6}{5}\)

Since \( 1+ \log (x) \) is an increasing function, and because \( \frac {4}{3} > \frac {5}{4} > \frac {6}{5} \)

Then \( \log_3 4 > \log_4 5 > \log_5 6 \) or \( \log_4 3 < \log_5 4 < \log_6 5 \) Pi Han Goh · 3 years, 9 months ago

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@Pi Han Goh Why do you need \(x>e\) ? Abhishek Sinha · 3 years, 9 months ago

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@Abhishek Sinha MY MISTAKE! TYPO. FIXED Pi Han Goh · 3 years, 9 months ago

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Is there any method using only algebra? Kishan K · 3 years, 9 months ago

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@Kishan K Perhaps this is the approach you are looking for: let \(a = \log_4(3)\); then \(0 < a < 1\) and \(4^a = 3\). So \[ 5^a = (4\cdot \frac{5}{4})^a = 4^a \cdot (5/4)^a = 3\cdot (5/4)^a < 3 \cdot 5/4 = 15/4 < 4, \] which implies that \(a < \log_5(4)\). So we showed that \(\log_4(3)<\log_5(4)\). A very similar method also shows that \(\log_5(4)<\log_6(5)\). John Smith Staff · 3 years, 9 months ago

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