Compare

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

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 - 7 years, 10 months ago

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

Abhishek Sinha - 7 years, 10 months ago

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

Kishan k - 7 years, 10 months ago

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 - 7 years, 10 months ago

@Pi Han Goh – Why do you need $x>e$ ?

Abhishek Sinha - 7 years, 10 months ago

@Abhishek Sinha – MY MISTAKE! TYPO. FIXED

Pi Han Goh - 7 years, 10 months ago

Is there any method using only algebra?

Kishan k - 7 years, 10 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 $\log_5(4)<\log_6(5)$ .

John Smith Staff - 7 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$