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Let's break down the problem using Lagrange's Mean Value Theorem (LMVT).

### Given Problem:
Using Lagrange's mean value theorem, prove that:

\[
\frac{b-a}{b} < \ln\left(\frac{b}{a}\right) < \frac{b-a}{a}
\]

for \(0 < a < b\). Hence, show that:

\[
\frac{1}{4} < \ln\left(\frac{4}{3}\right) < \frac{1}{3}
\]

### Step 1: Apply Lagrange’s Mean Value Theorem

Lagrange's Mean Value Theorem states that if a function \(f(x)\) is continuous on a closed interval \([a,b]\) and differentiable on the open interval \((a,b)\), then there exists a point \(c \in (a, b)\) such that:

\[
f'(c) = \frac{f(b) - f(a)}{b - a}
\]

Here, consider the function \(f(x) = \ln(x)\). This function is continuous and differentiable on \((0, \infty)\). Applying LMVT to \(f(x) = \ln(x)\) over the interval \([a, b]\), we get:

\[
f'(c) = \frac{\ln(b) - \ln(a)}{b - a}
\]

Now, the derivative of \(\ln(x)\) is:

\[
f'(x) = \frac{1}{x}
\]

Therefore, at some point \(c\) in \((a, b)\), we have:

\[
\frac{1}{c} = \frac

Using Lagrange's mean value theorem, prove that (b-a)/b < ln(b/a) < (b-a)/a for 0 < a < b. Hence, show that 1/4 < ln(4/3) < 1/3.

This problem involves using Lagrange's Mean Value Theorem to prove inequalities involving logarithmic functions. It also includes a demonstration of how to show 1/4 < ln(4/3) < 1/3 for 0 < a < b.

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