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We are asked to express \( \log 40 \) in terms of \( p = \log 5 \) and \( q = \log 2 \).

### Solution:
First, we express \( 40 \) as a product of prime factors:
\[
40 = 2^3 \times 5
\]
Now, applying the logarithmic property \( \log(ab) = \log a + \log b \), we get:
\[
\log 40 = \log(2^3 \times 5) = \log 2^3 + \log 5
\]

Using the power rule of logarithms \( \log(a^b) = b \log a \), this becomes:
\[
\log 40 = 3 \log 2 + \log 5
\]

Now substitute \( \log 5 = p \) and \( \log 2 = q \):
\[
\log 40 = 3q + p
\]

Thus, \( \log 40 \) in terms of \( p \) and \( q \) is:
\[
\log 40 = 3q + p
\]

Would you like further explanation or have any questions about this solution?

### Related Questions:
1. How would you express \( \log 100 \) in terms of \( p \) and \( q \)?
2. How does the change of base formula apply when rewriting logarithmic expressions?
3. Can you express \( \log 25 \) in terms of \( p \) and \( q \)?
4. How would you handle logarithms of non-integer values in terms of \( p \) and \( q \)?
5. How would you simplify \( \log(10) \) in terms of \( p \) and \( q \)?

### Tip:
Remember the key logarithmic properties: product rule \( \log(ab) = \log a + \log b \), quotient rule \( \log(a/b) = \log a - \log b \), and power rule \( \log(a^b) = b \log a \).

Given that p = log 5 and q = log 2, write log 40 in terms of p and q.

Learn how to express log 40 in terms of p = log 5 and q = log 2 using logarithmic properties. This solution demonstrates the use of the product and power rules of logarithms for students in Grades 10-12.

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