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.

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 \(\log_5(4)<\log_6(5)\).

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## Comments

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TopNewestLemma: 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 \( x \geq 2 \).

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Show that \(\frac{\log(x)}{\log(x+1)}\) is an increasing function for \(x>0\), where the base is e.

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I dont know much Calculus....please write ur solution.Is there any simple method......?

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

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Is there any method using only algebra?

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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 \(\log_5(4)<\log_6(5)\).

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