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

## Comments

Sort by:

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 \( x \geq 2 \).

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

Is there any method using only algebra?

Log in to reply

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