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The given equation is:

\[
9 + \log_3(-3b) = 8
\]

### Step 1: Isolate the logarithmic term.
Subtract 9 from both sides:

\[
\log_3(-3b) = 8 - 9
\]

\[
\log_3(-3b) = -1
\]

### Step 2: Convert the logarithmic equation to exponential form.
The equation \(\log_3(x) = y\) can be rewritten as \(x = 3^y\). Applying this to \(\log_3(-3b) = -1\):

\[
-3b = 3^{-1}
\]

\[
-3b = \frac{1}{3}
\]

### Step 3: Solve for \(b\).
Now, divide both sides by \(-3\) to isolate \(b\):

\[
b = \frac{\frac{1}{3}}{-3} = \frac{1}{3} \times \frac{-1}{3}
\]

\[
b = -\frac{1}{9}
\]

Thus, the solution for \(b\) is:

\[
b = -\frac{1}{9}
\]

Let me know if you'd like further clarification! Here are some related questions to practice:

1. What is the general method to convert a logarithmic equation to an exponential equation?
2. How do you handle negative bases inside a logarithmic function?
3. Can you apply logarithms to negative numbers? Why or why not?
4. What are the properties of logarithms that simplify solving logarithmic equations?
5. How would you solve an equation with a logarithm of a different base (e.g., base 10 or base 2)?

**Tip:** Always check the domain of logarithmic functions, as logarithms are only defined for positive arguments.

Solve for the variable: 9 + log_3(-3b) = 8

Learn how to solve the logarithmic equation 9 + log_3(-3b) = 8 by converting to exponential form. This problem covers the relationship between logarithms and exponents and algebraic manipulation. Ideal for students in Grades 9-12.

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