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.

This page provides a detailed step-by-step solution to the expression 2²log16^-1. The solution covers logarithmic properties, exponentiation rules, and simplification techniques. Ideal for high school students in Grades 10-12 looking to master these mathematical concepts.

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