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The problem asks us to find the linear approximation of \( f(x) = \ln(x) \) at \( x = 1 \) and then use it to estimate \( \ln(1.46) \).

### Step 1: Linear Approximation Formula

The linear approximation of a function \( f(x) \) around a point \( x = a \) is given by:

\[
L(x) = f(a) + f'(a)(x - a)
\]

In this case, \( f(x) = \ln(x) \) and \( a = 1 \).

### Step 2: Calculate \( f(1) \)

Since \( f(x) = \ln(x) \), we have:

\[
f(1) = \ln(1) = 0
\]

### Step 3: Calculate \( f'(x) \)

The derivative of \( f(x) = \ln(x) \) is:

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

Thus, \( f'(1) = \frac{1}{1} = 1 \).

### Step 4: Form the Linear Approximation \( L(x) \)

Using the formula \( L(x) = f(a) + f'(a)(x - a) \), we substitute \( a = 1 \), \( f(1) = 0 \), and \( f'(1) = 1 \):

\[
L(x) = 0 + 1(x - 1) = x - 1
\]

Thus, the linear approximation is:

\[
L(x) = x - 1
\]

### Step 5: Estimate \( \ln(1.46) \)

To estimate \( \ln(1.46) \), we substitute \( x = 1.46 \) into the linear approximation \( L(x) \):

\[
L(1.46) = 1.46 - 1 = 0.46
\]

Thus, the estimated value of \( \ln(1.46) \) is approximately \( 0.46 \).

### Final Answers:
1. The linear approximation \( L(x) = x - 1 \).
2. The estimated value of \( \ln(1.46) \approx 0.46 \).

Would you like further details or have any questions?

Here are 5 related questions for further exploration:
1. What is the error between the actual value of \( \ln(1.46) \) and the linear approximation?
2. How does linear approximation change if we expand around a different point, say \( x = 2 \)?
3. Why does linear approximation work well for values of \( x \) close to 1?
4. How does the linear approximation compare with higher-order approximations (e.g., Taylor series)?
5. What is the second derivative of \( \ln(x) \) and how does it impact the curvature near \( x = 1 \)?

**Tip:** Linear approximation is only accurate for values close to the point of approximation. For better accuracy, consider using more terms in the Taylor series.

Find the linear approximation of f(x) = ln(x) at x = 1 and use it to estimate ln(1.46).

Learn how to find the linear approximation of ln(x) at x = 1 using calculus and use this to estimate ln(1.46). This problem covers linear approximation formulas and derivative concepts, suitable for high school calculus students.

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