For \( g(x) = 2\ln x – \ln d \), rewrite using logarithmic properties:

\[ g(x) = \ln(x^2) – \ln d = \ln\left(\frac{x^2}{d}\right) \]

The vertical asymptote occurs when \( \frac{x^2}{d} = 0 \), so \( x^2 = 0 \), \( x = 0 \). As \( x \to 0^+ \), \( g(x) \to -\infty \).

Answer: \( x = 0 \)

(i) Set \( f(x) = g(x) \):

\[ \ln(2x – 9) = 2\ln x – \ln d \]

Rewrite: \( \ln(2x – 9) = \ln\left(\frac{x^2}{d}\right) \).

\[ 2x – 9 = \frac{x^2}{d} \]

Multiply by \( d \): \( d(2x – 9) = x^2 \).

\[ x^2 – 2dx + 9d = 0 \]

(ii) For two real roots, discriminant of \( x^2 – 2dx + 9d = 0 \):

\[ \Delta = (-2d)^2 – 4 \cdot 1 \cdot 9d = 4d^2 – 36d \]

\[ 4d(d – 9) > 0 \]

Since \( d > 0 \), \( d > 9 \). Thus, \( d^2 – 9d > 0 \).

(iii) From (ii), \( d > 9 \).

Answer: \( d > 9 \)

For \( d = 10 \), solve: \( \ln(2x – 9) = 2\ln x – \ln 10 \).

\[ \ln(2x – 9) = \ln\left(\frac{x^2}{10}\right) \]

\[ 2x – 9 = \frac{x^2}{10} \]

\[ x^2 – 20x + 90 = 0 \]

Discriminant: \( \Delta = (-20)^2 – 4 \cdot 1 \cdot 90 = 400 – 360 = 40 \).

\[ x = \frac{20 \pm \sqrt{40}}{2} = 10 \pm \sqrt{10} \]

\[ p = 10 – \sqrt{10}, \quad q = 10 + \sqrt{10} \]

\[ q – p = (10 + \sqrt{10}) – (10 – \sqrt{10}) = 2\sqrt{10} \]

Answer: \( 2\sqrt{10} \)