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

Note by Kishan K
3 years, 2 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 $$x \geq 2$$. Staff · 3 years, 2 months ago

Show that $$\frac{\log(x)}{\log(x+1)}$$ is an increasing function for $$x>0$$, where the base is e. · 3 years, 2 months ago

I dont know much Calculus....please write ur solution.Is there any simple method......? · 3 years, 2 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 $$\log_4 3 < \log_5 4 < \log_6 5$$ · 3 years, 2 months ago

Why do you need $$x>e$$ ? · 3 years, 2 months ago

MY MISTAKE! TYPO. FIXED · 3 years, 2 months ago

Is there any method using only algebra? · 3 years, 2 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 $$\log_5(4)<\log_6(5)$$. Staff · 3 years, 2 months ago