The expression in the image is:

$\frac{(4^{2k})^{-2} \times 4^{5k}}{4^{-k}}$

and the task is to solve for $k$.

Let's simplify this expression step by step.

Step 1: Simplify $(4^{2k})^{-2}$

Using the property $(a^m)^n = a^{m \cdot n}$, we can rewrite $(4^{2k})^{-2}$ as: $4^{2k \cdot (-2)} = 4^{-4k}$

So the expression becomes: $\frac{4^{-4k} \times 4^{5k}}{4^{-k}}$

Step 2: Combine the terms in the numerator

Since we are multiplying powers of 4, we can add the exponents: $4^{-4k} \times 4^{5k} = 4^{(-4k + 5k)} = 4^{k}$

Now the expression simplifies to: $\frac{4^{k}}{4^{-k}}$

Step 3: Simplify the division

Using the property $\frac{a^m}{a^n} = a^{m - n}$, we get: $4^{k - (-k)} = 4^{k + k} = 4^{2k}$

Step 4: Solve for $k$

Since we want to solve for $k$, we assume that the simplified expression equals a specific value, depending on any additional context or conditions. If it’s implied that $4^{2k} = 1$ or another specific target, we would set up an equation based on that. However, with no additional conditions, the simplified result is:

$4^{2k}$

If you need further steps based on a specific equation, let me know!

Would you like to delve deeper into any of these steps, or do you have other questions? Here are some questions to consider:

1. How do properties of exponents apply in other similar problems?
2. What if the base was not 4 but another number? How would that affect the solution?
3. How could logarithms be used to solve for $k$ if the final expression equals a specific number?
4. Can you solve similar expressions with variables in both the base and exponent?
5. How does this solution change if additional constraints on $k$ are introduced?

Tip: Always remember that exponent rules can simplify complex expressions significantly, especially in fraction-based problems.