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.

Is there any method using only algebra?
–
Kishan K
·
3 years, 5 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, 5 months ago

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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 \( x \geq 2 \). – Calvin Lin Staff · 3 years, 5 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, 5 months ago

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– Kishan K · 3 years, 5 months ago

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

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, 5 months ago

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– Abhishek Sinha · 3 years, 5 months ago

Why do you need \(x>e\) ?Log in to reply

– Pi Han Goh · 3 years, 5 months ago

MY MISTAKE! TYPO. FIXEDLog in to reply

Is there any method using only algebra? – Kishan K · 3 years, 5 months ago

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– John Smith Staff · 3 years, 5 months ago

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 \(\log_5(4)<\log_6(5)\).Log in to reply