If $\log(a) - \log(b) \gt \log(c) - \log(d)$, must it always follow that $(a-b) \gt (c-d)$?

I'm looking over someone else's work, and they inexplicably took the logarithms of the numbers and then compared them, instead of just comparing the numbers themselves. He ends up throwing them into a sorted list and separates them out into quartiles for further analysis.

This got me to wondering, would the quartiles contain the exact same entities if he had never taken the logarithms in the first place?

Unfortunately, I can't see the underlying data, so I can't just figure it out for myself. Also, I'm just a law student checking a faculty member's work, so I want to make sure I know what I'm talking about before I suggest a change to how he describes his results.


This is equivalent to $$\log (a/b) > \log (c/d)$$

But the logarithm function is increasing, so this is equivalent to $$a/b > c/d$$

Does this imply $a - b > c-d $? Not necessarily. You can set $a = d = 2, c = 3, b = 1$. Then $a/b = 2 > 3/2 = c/d$, but $a - b = 1 = c-d$.

Explicitly, we observe that $\log 2 - \log 1 \sim 0.301 > 0.176 \sim \log 3 - \log 2$, but of course $a - b = c-d$.


$log(a/b) > log(c/d)$

$\implies a/b > c/d$

$\implies (a-b)/b > (c-d)/d$

$\implies a-b > (c-d)b/d$

$\implies$ For $b \geq d$, $a-b > c-d$


Err, the person that wrote this is a faculty member? In a Mathematics department? This is trivially provable to be false.

Let $a = 2, b = 1, c = 3, d = 2$, then:

$a - b = 2 - 1 = 1$ and $c - d = 3 - 2 = 1$

Then although, $log(2) - log(1) > log(3) - log(2)$ we have that $1 \not \gt 1$

  • $\begingroup$ No, the law professor's math is right. I just couldn't figure out why he had to take the differences of the natural logs of two set of numbers when he was just going to rank the difference in a particular order. In other words, I was hoping he could simplify the explanation in his footnotes by removing what had appeared to me to be an unnecessary step. And this is why I'm really just supposed to be editing for style and not substance. Thanks. $\endgroup$ – Shane Mar 22 '13 at 4:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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