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.