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 $\log(a^b) = b \log a$.