(a) Finding Translation Values:

Start with \( y = \ln x \) and apply transformations:
Step 1: Horizontal translation \( a \) units: \( y = \ln(x – a) \)
Step 2: Vertical translation \( b \) units: \( y = \ln(x – a) + b \)

Compare with given function:
\( \ln(x – a) + b = \ln(5x + 10) \)
Rewrite right side: \( \ln(5(x + 2)) = \ln 5 + \ln(x + 2) \)

Therefore:
\( \ln(x – a) + b = \ln 5 + \ln(x + 2) \)
This gives:
\( x – a = x + 2 \) ⇒ \( \boxed{a = -2} \)
\( b = \ln 5 \) ⇒ \( \boxed{b = \ln 5} \) (≈1.609)

(b) Volume of Revolution:

Volume formula: \( V = \pi \int_{e}^{2e} [\ln(5x + 10)]^2 dx \)

Substitution: Let \( u = 5x + 10 \) ⇒ \( du = 5dx \) ⇒ \( dx = \frac{du}{5} \)
Limits change:
When \( x = e \), \( u = 5e + 10 \)
When \( x = 2e \), \( u = 10e + 10 \)

Integral becomes:
\( V = \frac{\pi}{5} \int_{5e+10}^{10e+10} (\ln u)^2 du \)

Integration by parts:
\( \int (\ln u)^2 du = u(\ln u)^2 – 2u\ln u + 2u + C \)

Evaluate at limits:
\( V = \frac{\pi}{5}[(10e+10)(\ln(10e+10))^2 – 2(10e+10)\ln(10e+10) + 2(10e+10) \)
\( – (5e+10)(\ln(5e+10))^2 + 2(5e+10)\ln(5e+10) – 2(5e+10)] \)

Numerical approximation:
\( \boxed{V \approx 99.2} \)