We are given that:

$\log 27 = 1.431$

Our task is to find the value of $\log 9$.

First, express both 27 and 9 in terms of powers of 3: $27 = 3^3 \quad \text{and} \quad 9 = 3^2$

Using the logarithmic identity: $\log(a^b) = b \cdot \log a$

We can express $\log 27$ and $\log 9$ as follows: $\log 27 = \log(3^3) = 3 \cdot \log 3$ $\log 9 = \log(3^2) = 2 \cdot \log 3$

From $\log 27 = 1.431$, we know: $3 \cdot \log 3 = 1.431$

Solving for $\log 3$: $\log 3 = \frac{1.431}{3} = 0.477$

Now, substitute $\log 3$ into the expression for $\log 9$: $\log 9 = 2 \cdot \log 3 = 2 \cdot 0.477 = 0.954$

Thus, the value of $\log 9$ is:

$\boxed{0.954}$

Answer: d. 0.954

Do you have any questions or want more details on the steps?

Here are 5 questions to further explore logarithms:

1. What is the logarithmic property that relates the logarithm of a product to individual logarithms?
2. How would you find $\log 81$ if you know $\log 27$?
3. What is the change of base formula for logarithms, and how is it used?
4. How does the value of $\log 9$ change if the base is 2 instead of 10?
5. Can you explain why logarithms are useful in solving exponential equations?

Tip: When working with logarithms, always try to express the numbers in terms of powers of the same base for simpler calculations.