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.