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I'm trying to prove

$1 - \frac{a}{b} \leq \ln\frac{b}{a}\leq\frac{a}{b} - 1$ where $0 < a < b$ using Lagrange's Mean Value Theorem.

Applying the theorem to $\ln x$ results in: $$\exists\epsilon\in(a,b): \ln'(\epsilon)(b-a)=\ln b - \ln a$$ $$\frac{b}{\epsilon}-\frac{a}{\epsilon}=\ln \frac{b}{a}$$

This looks very similar to the target inequality (set $\epsilon=a$ and $\epsilon=b$), but I'm not sure how to get to it.

Edit: Looks like my question is an exact duplicate of Mean Value theorem problem?(inequality), The answer doesn't really explain how to get to the inequalities though.


marked as duplicate by Martin R, Robert Z, Alessandro Codenotti, Clarinetist, Fabian Jan 5 '18 at 17:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Looks like a clear duplicate to me. What are you missing in math.stackexchange.com/a/618533/42969? $\endgroup$ – Martin R Jan 5 '18 at 14:40
  • $\begingroup$ @MartinR How they get to the inequalities: why is $\frac{b-a}{b} < \ln b - \ln a$? $\endgroup$ – Todd Sewell Jan 5 '18 at 14:41
  • $\begingroup$ $ \ln b - \ln a = \frac{b-a}{c} > \frac{b-a}{b}$ because $a < c < b$. $\endgroup$ – Martin R Jan 5 '18 at 14:45
  • $\begingroup$ @MartinR Yes, I understand it now, thanks to both you and the answers I got on this question. $\endgroup$ – Todd Sewell Jan 5 '18 at 14:47
  • $\begingroup$ @ToddSewell If you are ok, you can accept the answer and set as solved. Thanks! $\endgroup$ – gimusi Jan 8 '18 at 22:40

Note that fro MVT

$$\ln b - \ln a=\ln \frac{b}a=\frac{b-a}{c} \quad c\in(a,b)$$

and varing $c$ between $a$ and $b$

$$1-\frac{a}{b}\leq\frac{b-a}{c}\leq \frac{b}{a}-1$$


$$1-\frac{a}{b}\leq\ln \frac{b}a\leq \frac{b}{a}-1$$


use that $$\frac{\ln(b)-\ln(a)}{b-a}=\frac{1}{\xi}$$ and $$\xi \in (a,b)$$ it is not difficult to complete this, since we have $$0<a<b$$ we get that $$\frac{1}{\xi}<\frac{1}{a}$$ and $$\frac{1}{\xi}>\frac{1}{b}$$

  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ – Arnaud D. Jan 5 '18 at 17:36
  • $\begingroup$ and why is it so like you say? $\endgroup$ – Dr. Sonnhard Graubner Jan 5 '18 at 17:39
  • $\begingroup$ OP already has that equality written in the question. The question is how to find the inequalities from that equality, and your answer doesn't really adress that. $\endgroup$ – Arnaud D. Jan 5 '18 at 17:42
  • $\begingroup$ now it does i think $\endgroup$ – Dr. Sonnhard Graubner Jan 5 '18 at 17:46

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