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I need to prove: $$\frac{a-b}{a} < \ln(\frac{a}{b}) < \frac{a-b}{b}$$ $$0<b<a$$

I tried the substitution $x = a/b$ and analyzing it as three functions, but it led me nowhere.
Any hints would be welcome.

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    $\begingroup$ Hint: Remember that $\ln(\frac{a}{b})=\ln(a)-\ln(b)$. $\endgroup$ – Michael Burr Feb 26 '16 at 14:58
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METHODOLOGY $1$:

I thought it might be instructive to present a way forward that does not rely on calculus.

In THIS ANSWER, I showed using only the limit definition of the logarithm and Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\frac{x-1}{x}\le \log(x)\le x-1$$

for $x>0$. Now, simply set $x=a/b$, where $0<b<a$, and we find

$$\frac{a-b}{a} \le \log(a/b)\le \frac{a-b}{b}$$

as expected.


METHODOLOGY $2$:

Here, we use the integral definition of the natural logarithm along with the Mean Value Theorem for integrals. Let $\log(x)$ be defined by

$$\log(x)\equiv \int_1^x \frac{1}{t}\,dt \tag 1$$

Then, setting $x=a/b$ in $(1)$, where $0<a<b$ reveals

$$\begin{align} \log(a/b)&=\int_1^{a/b}\frac{1}{t}\,dt\\\\ &=-\int_{a/b}^1\frac{1}{t}\,dt \tag 2\\\\ \end{align}$$

Using the Mean Value Theorem, there exists a number $\xi\in(a/b,1)$ such that

$$-\int_{a/b}^1\frac{1}{t}\,dt=-\frac{1}{\xi}\left(1-\frac{a}{b}\right)=\frac{a-b}{\xi \, b} \tag 3$$

Since $a/b<\xi<1$, then we have immediately from $(3)$ that

$$\frac{a-b}{a}\le \log(a/b)<\frac{a-b}{b}$$

as was to be shown!

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    $\begingroup$ I love chess too! e4 ... - Mark $\endgroup$ – Mark Viola Feb 26 '16 at 20:14
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Hint: consider $f(x) =\ln x$ on $(b,a)$

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For $x \in [b, a] \quad \frac{1}{a} \leq \frac{1}{x} \leq \frac{1}{b} $. hence $\int_b^{a} \frac{dx}{a} \leq \int_b^{a} \frac{dx}{x} \leq \int_b^{a} \frac{dx}{b} \iff \frac{a-b}{a} \leq Ln(\frac{a}{b} ) \leq \frac{a-b}{b} $. Strict inequalities: for instance $\int_b^{a} (\frac{1}{b} - \frac{1}{x} ) dx > 0$ for we integrate a strictly positive continuous function over $[b, a]$ (with $b < a$).

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Principle: On $[a,b]$, $(b > a)$, we have $(b-a)\min_{u \in [a,b]} f(u) \le \int_a^b f(u) du \le \max_{u \in [a,b]} f(u).$

We have $\ln \frac{b}{a}=\ln(b) - \ln(a) = \int_a^b \frac1{u} du.$

but $1/b < 1/u < 1/a$ so $\frac{(b-a)}{b} < \ln(b)-\ln(a) < \frac{(b-a)}{a}.$

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