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Let's solve the questions step by step.

### The given curve is:  
\[ y = \ln(3x - 1) \]
and point \( P \) is where \( x = 2 \).

---

### (a) **Gradient of the curve at \( P \)**

To find the gradient of the curve at \( P \), we need to differentiate \( y \) with respect to \( x \) and then evaluate it at \( x = 2 \).

1. **Differentiate \( y = \ln(3x - 1) \):**

Using the chain rule:
\[
\frac{dy}{dx} = \frac{1}{3x - 1} \cdot \frac{d}{dx}(3x - 1) = \frac{1}{3x - 1} \cdot 3 = \frac{3}{3x - 1}
\]

2. **Evaluate at \( x = 2 \):**
\[
\frac{dy}{dx} \Big|_{x=2} = \frac{3}{3(2) - 1} = \frac{3}{6 - 1} = \frac{3}{5}
\]

Thus, the gradient at \( P \) is:
\[
\boxed{\frac{3}{5}}
\]

---

### (b) **Equation of the tangent to the curve at \( P \)**

The equation of the tangent line is given by the point-slope form:
\[
y - y_1 = m(x - x_1)
\]
where \( m \) is the gradient, and \( (x_1, y_1) \) is the point of tangency.

1. From part (a), the gradient \( m = \frac{3}{5} \).

2. The point \( P \) is where \( x = 2 \). Let's find \( y \) at \( x = 2 \):
\[
y = \ln(3(2) - 1) = \ln(6 - 1) = \ln(5)
\]
So, the coordinates of \( P \) are \( (2, \ln(5)) \).

3. Substitute into the equation:
\[
y - \ln(5) = \frac{3}{5}(x - 2)
\]
Simplify:
\[
y = \frac{3}{5}(x - 2) + \ln(5)
\]

Thus, the equation of the tangent is:
\[
\boxed{y = \frac{3}{5}x - \frac{6}{5} + \ln(5)}
\]

---

### (c) **Coordinates of \( R \) where the normal to the curve at \( P \) cuts the \( x \)-axis**

1. The gradient of the normal is the negative reciprocal of the gradient of the tangent. So, the gradient of the normal is:
\[
m_{\text{normal}} = -\frac{5}{3}
\]

2. The equation of the normal is also given by the point-slope form:
\[
y - \ln(5) = -\frac{5}{3}(x - 2)
\]
Simplify:
\[
y = -\frac{5}{3}(x - 2) + \ln(5)
\]
\[
y = -\frac{5}{3}x + \frac{10}{3} + \ln(5)
\]

3. To find the x-intercept, set \( y = 0 \):
\[
0 = -\frac{5}{3}x + \frac{10}{3} + \ln(5)
\]
Multiply through by 3 to eliminate the fractions:
\[
0 = -5x + 10 + 3\ln(5)
\]
Solve for \( x \):
\[
5x = 10 + 3\ln(5)
\]
\[
x = 2 + \frac{3\ln(5)}{5}
\]

Thus, the x-coordinate of \( R \) is:
\[
\boxed{x = 2 + \frac{3\ln(5)}{5}}
\]

The coordinates of \( R \) are \( \left(2 + \frac{3\ln(5)}{5}, 0\right) \).

---

Do you want further details or have any questions?  
Here are some related questions to expand on this:

1. How do you find the second derivative of \( y = \ln(3x - 1) \)?
2. Can you find the point where the curve has zero gradient?
3. What is the area under the curve between \( x = 1 \) and \( x = 2 \)?
4. How would the equation of the tangent change if the curve was \( y = \ln(2x - 3) \)?
5. What is the significance of the normal line's x-intercept in this problem?

**Tip**: When dealing with logarithmic functions, remember that their derivatives follow the chain rule and are often simplified using properties of logarithms.

Consider the curve y = ln(3x - 1). Let P be the point on the curve where x = 2. (a) Write down the gradient of the curve at P. (b) Find the equation of the tangent to the curve at P. (c) The normal to the curve at P cuts the x-axis at R. Find the coordinates of R.

This page provides a step-by-step solution to find the gradient, equation of the tangent, and the coordinates of the normal's x-intercept for the curve y = ln(3x - 1) at x = 2. Concepts include differentiation, tangent and normal lines, and logarithmic functions.

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