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).

Learn how to solve the logarithmic equation log_5(3x + 40) = 2 by converting it to exponential form and solving for x. This step-by-step solution covers key algebraic and logarithmic concepts, making it suitable for students in Grades 9-11.

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