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This question already has an answer here:

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.

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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
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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 $b$

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

thus

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

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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 $$\frac{1}{\xi}>\frac{1}{b}$$

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  • $\begingroup$ 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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