Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I realize that I lack any numerical intuition for logarithms. For example, when comparing two logarithms like $\log_78$ and $\log_89$, I have to use the change-of-base formula and calculate the values with a calculator in order to determine which of the two is larger.

Can anyone demonstrate an algebraic/analytic method to find out which of the two is larger without using a calculator?

share|improve this question
1  
$\log_x(x+1)=\log(x+1)/\log(x)$ is decreasing for $x>1$ but I don't know how to prove it... –  lhf May 3 '12 at 10:50
5  
@lhf: $\frac{\log(x + 2) / \log (x + 1)}{\log(x + 1) / \log(x)} = \frac{\log(x + 2) \log(x)}{\log(x + 1)^2}$. Now $\frac{(x + 2)x}{(x + 1)^2} < 1$ (rectangle areas), and since $\log$ is concave, the same must hold for when you apply $\log$ to each factor. –  Alexei Averchenko May 3 '12 at 12:22

4 Answers 4

up vote 26 down vote accepted

$\log_7 8 = 1 + \log_7 (8 / 7) > 1 + \log_7 (9/8) > 1 + \log_8(9/8) = \log_8 9$

share|improve this answer
    
brilliant! Thanks. –  hollow7 May 3 '12 at 11:23

The derivative of $f(x)=\frac{\log(x+1)}{\log(x)}$ has the same sign as $\frac{\log x}{x+1}-\frac{\log(x+1)}{x}$ which is negative if $x>1$ since $x\mapsto x\log{x}$ is increasing.

Of course, this method does not apply for arbitrary 7,8,9. For example $\log_35$ and $\log_23$ are quite close and proving that one is bigger than the other is not so easy. The only elegant way I know is some kind of trick. (Enjoy this entertaining exercise! Spoiler below.)

Prove that $\log_35 < \frac32 < \log_23$.

share|improve this answer
    
I don't get why the two expressions have the same sign. But actually I'm looking for a general method to compare the sizes of two logarithms. So the numbers 7,8,9 in the question are actually arbitrary. –  hollow7 May 3 '12 at 11:03
    
@dragoncharmer: $f'(x)\big(\log x\big)^2=\frac{\log x}{x+1}-\frac{\log(x+1)}x$, and $\big(\log x\big)^2>0$ for $x\ne 1$. –  Brian M. Scott May 3 '12 at 11:11
    
interesting exercise. it seems to me wallace's method fails here. –  hollow7 May 3 '12 at 11:35
    
btw in the exercise, 1/2 should be 3/2. –  hollow7 May 3 '12 at 11:43
    
@dragoncharmer: oh, thanks. –  jmad May 3 '12 at 12:07

Alternate solution:

$$\log_78 > \log_89 \Leftrightarrow \frac{1}{\log_87} > \log_89 \Leftrightarrow 1> \log_87 \log_89 $$

Now, by AM-GM

$$\log_87 \log_89 \leq (\frac{\log_87+ \log_89}{2})^2=(\frac{\log_863}{2})^2< (\frac{\log_864}{2})^2=1$$

In general If $ab < c^2$ and $\log_b c >0$ you have

$$\log_ca \log_cb \leq (\frac{\log_ca+ \log_cb}{2})^2< 1$$

and thus

$$\log_ca < \log_bc \,.$$

share|improve this answer

Consider the function $ f(x)=log_x(x+1)$

This is a decreasing function, hence ,

$$log_n(n+1) >log_m(m+1) $$ for $n < m$

hence , $$log_78 > log_89$$


f(x) is decreasing because, f '(x) comes out to be negative,

$$f(x)=\frac{ln(x+1)}{ln(x)}$$

$$f '(x)=\frac{1}{(x+1)lnx} + ln(x+1).\frac{-1}{{(lnx)}^2}.\frac{1}{x}$$

$$f '(x)=\frac{1}{lnx}(\frac{1}{x+1}-\frac{f(x)}{x})$$

since, $f(x)<1$ for $x \in R+$

hence$f'(x)<0$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.