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We are given the following problem: 

- A tangent line \( \overline{AB} \) is drawn to the curve \( y = \ln\left(\frac{x}{2}\right) \) at the point \( C(1, y) \).
- The line intersects the x-axis at point \( A \) and the y-axis at point \( B \).
- We are tasked with finding the length of the tangent line \( \overline{AB} \).

### Steps to solve:

1. **Find the derivative of the function** \( y = \ln\left(\frac{x}{2}\right) \).
   \[
   \frac{dy}{dx} = \frac{1}{x}
   \]
   This gives us the slope of the tangent line at any point on the curve.

2. **Find the slope of the tangent line at point \( C(1, y) \)**:
   At \( C(1, y) \), the slope of the tangent line is:
   \[
   \frac{dy}{dx}\Big|_{x=1} = \frac{1}{1} = 1
   \]
   So, the slope of the tangent line at point \( C(1, 0) \) is 1.

3. **Equation of the tangent line**:
   The equation of the tangent line in point-slope form is:
   \[
   y - y_1 = m(x - x_1)
   \]
   Substituting \( m = 1 \) and \( (x_1, y_1) = (1, \ln\left(\frac{1}{2}\right)) = (1, -\ln 2) \):
   \[
   y - (-\ln 2) = 1(x - 1)
   \]
   \[
   y + \ln 2 = x - 1
   \]
   \[
   y = x - 1 - \ln 2
   \]
   This is the equation of the tangent line.

4. **Find the x-intercept (point A)**:
   To find the x-intercept, set \( y = 0 \) and solve for \( x \):
   \[
   0 = x - 1 - \ln 2
   \]
   \[
   x = 1 + \ln 2
   \]
   So, point \( A \) is \( (1 + \ln 2, 0) \).

5. **Find the y-intercept (point B)**:
   To find the y-intercept, set \( x = 0 \) and solve for \( y \):
   \[
   y = 0 - 1 - \ln 2
   \]
   \[
   y = -1 - \ln 2
   \]
   So, point \( B \) is \( (0, -1 - \ln 2) \).

6. **Calculate the length of \( \overline{AB} \)**:
   The length of the line segment \( \overline{AB} \) can be calculated using the distance formula:
   \[
   \text{Length of } \overline{AB} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
   \]
   Substituting the coordinates of points \( A(1 + \ln 2, 0) \) and \( B(0, -1 - \ln 2) \):
   \[
   \text{Length of } \overline{AB} = \sqrt{(0 - (1 + \ln 2))^2 + ((-1 - \ln 2) - 0)^2}
   \]
   \[
   = \sqrt{(1 + \ln 2)^2 + (-1 - \ln 2)^2}
   \]
   \[
   = \sqrt{(1 + \ln 2)^2 + (1 + \ln 2)^2}
   \]
   \[
   = \sqrt{2(1 + \ln 2)^2}
   \]
   \[
   = \sqrt{2}(1 + \ln 2)
   \]
   Hence, the length of \( \overline{AB} \) is:
   \[
   \boxed{\sqrt{2}(1 + \ln 2) \, \text{units}}
   \]

Would you like further clarification on any step? Here are five related questions:

1. How would the equation of the tangent line change if the point of tangency were different?
2. Can we find the length of \( \overline{AB} \) using a different method?
3. How do we interpret the natural logarithmic function geometrically on this curve?
4. What happens to the slope of the tangent line as \( x \to \infty \)?
5. What role does the point \( C(1, y) \) play in determining the tangent's properties?

**Tip**: Always differentiate the function first when dealing with tangent lines to a curve!

If AB is a tangent to the curve y = ln(x/2) at point C(1, y) and intersects X-axis at point A and Y-axis at point B, find the length of AB.

This problem involves finding the length of a tangent line to the curve y = ln(x/2), using differentiation and the distance formula. A step-by-step solution helps students understand tangent lines and logarithmic derivatives. Suitable for Grades 11-12.

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