Prove an inequality using Mean Value Theorem 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.
 A: Hint: consider $f(x) =\ln x$ on $(b,a)$
A: 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 $b < a$).
A: 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 $\frac{(b-a)}{b} < \ln(b)-\ln(a) < \frac{(b-a)}{a}.$ 
A: 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!
