$$ 3^{2x+1}=\left(\frac{1}{27}\right)^{x+8} $$

We are given the equation:

\[
3^{2x+1} = \left(\frac{1}{27}\right)^{x+8}
\]

### Step 1: Rewrite the equation with powers of 3
Recall that $ \frac{1}{27} = 3^{-3} $. So, we can rewrite the equation as:

\[
3^{2x+1} = (3^{-3})^{x+8}
\]

### Step 2: Simplify the right-hand side
Use the power rule $ (a^m)^n = a^{m \cdot n} $ to simplify the right-hand side:

\[
3^{2x+1} = 3^{-3(x+8)}
\]

Simplifying the exponent on the right-hand side:

\[
3^{2x+1} = 3^{-3x - 24}
\]

### Step 3: Set the exponents equal
Since the bases are the same (both base 3), we can set the exponents equal to each other:

\[
2x + 1 = -3x - 24
\]

### Step 4: Solve for $ x $
First, add $ 3x $ to both sides to combine like terms:

\[
5x + 1 = -24
\]

Now, subtract 1 from both sides:

\[
5x = -25
\]

Finally, divide by 5:

\[
x = -5
\]

### Final Answer:
\[
x = -5
\]

Would you like more details or have any questions?

Here are 5 related questions:
1. What if the equation was $ 3^{2x+1} = 27^{x+8} $?
2. How would the solution change if the base on the right-hand side was 9 instead of 27?
3. Can you solve $ 3^{x} = \frac{1}{81} $?
4. What would happen if the equation was $ 3^{2x+1} = \frac{1}{3^{x+8}} $?
5. What happens if you multiply both sides by 3 in the original equation?

**Tip:** Always check if the bases in the equation can be expressed as powers of a common base to simplify the process.

Solve the equation: 3^{2x+1} = \left(\frac{1}{27}\right)^{x+8}

Learn how to solve the exponential equation 3^{2x+1} = (1/27)^{x+8} using exponent rules and algebraic manipulation. This solution explains how to express both sides with a common base and solve for x. Ideal for students in Grades 9-11.

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