To solve the expression $2^{2 \log 16^{-1}}$, follow these steps:

Step 1: Simplify the logarithm expression

We start by simplifying the logarithm: $\log 16^{-1} = -\log 16$

Next, express 16 as a power of 2: $\log 16 = \log 2^4 = 4 \log 2$ So: $\log 16^{-1} = -4 \log 2$

Step 2: Substitute and simplify

Now, substitute back into the original expression: $2^{2 \log 16^{-1}} = 2^{2(-4 \log 2)} = 2^{-8 \log 2}$

Step 3: Apply the exponent rule

Use the exponentiation rule $a^{\log_b a} = b$: $2^{-8 \log 2} = 2^{\log 2^{-8}} = 2^{-8}$

Step 4: Simplify the final expression

Finally, simplify $2^{-8}$: $2^{-8} = \frac{1}{2^8} = \frac{1}{256}$

Thus, the value of the expression is $\boxed{\frac{1}{256}}$.

Would you like more details or have any questions about this process?

Here are 5 related questions to expand your understanding:

1. How do logarithm properties simplify expressions like $\log 2^4$?
2. What is the significance of the base in logarithmic expressions?
3. How do negative exponents affect the value of an expression?
4. Can you express $16^{-1}$ in terms of a fraction before applying the logarithm?
5. What is the general form of the exponentiation rule used in solving $a^{\log_b a}$?

Tip: When dealing with logarithms, always try to express the number inside the log as a power of the base, as this simplifies the calculation.