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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?

  • 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.2012-05-03

4 Answers 4

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

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    brilliant! Thanks.2012-05-03
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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 $\log_35 < \frac32 < \log_23$.

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    I think I found a more direct proof, see my comment to the question.2012-05-03
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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

$\log_ca < \log_bc \,.$

2

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

This is a decreasing function and hence $\log_n(n+1) >\log_m(m+1)$ for $n < m$ and hence $\log_78 > \log_89$.


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

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

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

$f '(x)=\frac{1}{\ln x}\left(\frac{1}{x+1}-\frac{f(x)}{x}\right)$

and since $f(x)<1$ for $x \in \mathbb R^+$, we have $f'(x)<0$.