Step 1: Understanding the Concept:
The normal to a curve at a point is a straight line perpendicular to the tangent at that point.
The slope of the tangent (\(m_T\)) at any point \((x, y)\) is given by the derivative \(dy/dx\).
The slope of the normal (\(m_N\)) is the negative reciprocal of the slope of the tangent: \(m_N = -1/m_T\).
Once the slope and point are known, we use the point-slope form to find the line's equation.
Step 2: Key Formula or Approach:
1. Curve: \(y = \ln x\).
2. Derivative: \(\frac{dy}{dx} = \frac{1}{x}\).
3. Point-slope form: \(y - y_1 = m(x - x_1)\).
Step 3: Detailed Explanation:
Differentiate the function \(y = \log_e x\):
\[ \frac{dy}{dx} = \frac{1}{x} \]
Find the slope of the tangent at point \(P(1, 0)\):
\[ m_T = \left[ \frac{1}{x} \right]_{x=1} = \frac{1}{1} = 1 \]
Since the normal is perpendicular to the tangent, its slope \(m_N\) is:
\[ m_N = -\frac{1}{m_T} = -\frac{1}{1} = -1 \]
Now, use the equation of a line with slope \(-1\) passing through \((1, 0)\):
\[ y - 0 = -1(x - 1) \]
\[ y = -x + 1 \]
\[ x + y = 1 \]
This is the linear equation representing the normal to the log curve at its x-intercept.
Step 4: Final Answer:
The equation of the normal is \(x + y = 1\).
Hence, the correct option is (D).