The problem asks to calculate $\log 18$ given $\log 2 = p$ and $\log 3 = q$. To find the value of $\log 18$, we use the properties of logarithms.

Solution:

1. Factorize 18:
   $18 = 2 \times 3^2$.

2. Apply the logarithmic property:
   $\log (a \times b) = \log a + \log b$, so
   $\log 18 = \log 2 + \log 3^2.$

3. Simplify $\log 3^2$:
   $\log 3^2 = 2 \log 3$.

   Therefore:
   $\log 18 = \log 2 + 2 \log 3.$

4. Substitute given values:
   $\log 2 = p$ and $\log 3 = q$, so
   $\log 18 = p + 2q.$

Answer:

The correct option is C. $p + 2q$.

Would you like further clarification or examples? Here are 5 related questions to expand your understanding:

1. How would the result change if $\log 5 = r$ and you calculate $\log 90$?
2. Can you prove $\log (ab) = \log a + \log b$ using logarithm properties?
3. What if $\log 3 = q$ but you are asked for $\log 27$?
4. How do you simplify $\log (2^3 \times 3^4)$ using the same principles?
5. What are the inverse operations of logarithms for equations involving $\log 18 = p + 2q$?

Tip:

To simplify logarithmic expressions, always try breaking down the number into its prime factors. This makes applying logarithmic properties much easier!