To solve the equation $2^{3x-8} = \frac{1}{4}$, follow these steps:

Step 1: Rewrite the equation in exponential form.

We know that $\frac{1}{4} = 2^{-2}$, so we can rewrite the equation as: $2^{3x-8} = 2^{-2}$

Step 2: Equate the exponents.

Since the bases are the same, the exponents must be equal: $3x - 8 = -2$

Step 3: Solve for $x$.

1. Add 8 to both sides: $3x = 6$

2. Divide both sides by 3: $x = 2$

Final Solution:

$x = 2$

Let me know if you'd like a detailed explanation of any step or have further questions!

Follow-Up Questions:

1. How would the solution change if the base was different (e.g., $3^{3x-8}$)?
2. Can you solve a similar equation $2^{3x - 5} = 8$?
3. What happens if the right-hand side of the equation was $0$?
4. How does this method work for non-integer bases, e.g., $e^{3x-8} = 1/e^2$?
5. Can this method be generalized for logarithmic equations?

Tip:

Always check whether the base and exponent format allow simplifications before proceeding with solving equations.