We are given the equation:

$8^{-2} = \frac{4}{4^{4x}}$

Step 1: Simplify the left-hand side

We know that $8^{-2} = \frac{1}{8^2} = \frac{1}{64}$. So, the equation becomes:

$\frac{1}{64} = \frac{4}{4^{4x}}$

Step 2: Simplify the right-hand side

We can rewrite the right-hand side as:

$\frac{4}{4^{4x}} = 4^{1 - 4x}$

So the equation now becomes:

$\frac{1}{64} = 4^{1 - 4x}$

Step 3: Express 64 as a power of 4

Since $64 = 4^3$, we can rewrite the left-hand side as:

$\frac{1}{64} = 4^{-3}$

Now the equation is:

$4^{-3} = 4^{1 - 4x}$

Step 4: Set the exponents equal to each other

Since the bases are the same, we can equate the exponents:

$-3 = 1 - 4x$

Step 5: Solve for $x$

Subtract 1 from both sides:

$-3 - 1 = -4x$

$-4 = -4x$

Now divide both sides by -4:

$x = 1$

Final Answer:

The value of $x$ is $1$.

Would you like more details or have any questions?

Here are some related questions to expand your understanding:

1. How do you solve exponential equations with different bases?
2. What is the general rule for negative exponents?
3. How do you simplify fractions involving exponents?
4. What happens when you equate exponents with the same base?
5. Can exponents ever be negative, and how do they affect the value?

Tip: Always try to express all terms with the same base when solving exponential equations.