To solve the equation $3^x = 31$, we can take the logarithm of both sides. Using the natural logarithm (ln), we have:

$\ln(3^x) = \ln(31)$

Using the power rule of logarithms:

$x \cdot \ln(3) = \ln(31)$

Now, solve for $x$:

$x = \frac{\ln(31)}{\ln(3)}$

Calculating this:

$x \approx \frac{3.43399}{1.09861} \approx 3.128$

So the solution is:

A. $x \approx 3.128$ (rounded to three decimal places).

If you need any further details or have questions, feel free to ask! Here are some related questions to consider:

1. What are the properties of logarithms used in solving exponential equations?
2. How can you check if the solution is correct?
3. What would happen if the base was a different number, such as 2 instead of 3?
4. How does this method apply to solving $a^x = b$ for other values of $a$ and $b$?
5. Can you explain how to graph the function $y = 3^x$ and where it intersects with $y = 31$?

Tip: When dealing with exponential equations, using logarithms is often the most effective approach for isolating the variable.