To solve the logarithmic equation:

$\log_5 (3x + 40) = 2$

we follow these steps:

Step 1: Rewrite the equation in exponential form.

The logarithmic equation $\log_b(y) = x$ can be rewritten as $b^x = y$. Using this property:

$5^2 = 3x + 40$

This simplifies to:

$25 = 3x + 40$

Step 2: Solve for $x$.

Now, subtract 40 from both sides:

$25 - 40 = 3x$

$-15 = 3x$

Now, divide both sides by 3:

$x = \frac{-15}{3}$

$x = -5$

Thus, the solution is:

$x = -5$

Would you like further clarification or more details?

Here are 5 related questions:

1. What is the general form of a logarithmic equation?
2. How do you convert a logarithmic equation into exponential form?
3. What happens when the base of a logarithm is not explicitly written?
4. Can you solve for $x$ if the base was a different number, like 10 or 2?
5. How would this equation change if there was an added constant on the left side?

Tip: Always check if the argument inside the logarithm remains positive after solving for $x$ (since logarithms are only defined for positive values).