To solve the equation \( \log_4 16 = x^x \), let's break it down into manageable steps.

### Step 1: Evaluate the logarithm
The logarithm \( \log_4 16 \) asks, "To what power must 4 be raised to get 16?" Since \( 4^2 = 16 \), we have:

\[
\log_4 16 = 2
\]

### Step 2: Substitute the logarithm value into the equation
Now that we know \( \log_4 16 = 2 \), substitute this into the equation:

\[
2 = x^x
\]

### Step 3: Solve for \( x \)
We now have the equation \( x^x = 2 \).

This type of equation doesn't have a simple algebraic solution, but we can solve it by inspection or numerically.

- If \( x = 2 \), then \( x^x = 2^2 = 4 \) (too large).
- If \( x = 1.5 \), then \( x^x \approx 1.5^{1.5} \approx 1.837 \) (still too small).
- If \( x = \sqrt{2} \), then \( x^x = (\sqrt{2})^{\sqrt{2}} = 2 \) exactly.

So, \( x = \sqrt{2} \) (approximately 1.414) is the solution.

### Final Answer:
The solution to the equation is:

\[
x = \sqrt{2}
\]

Would you like more details on any step or have any questions?

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Here are 5 related questions to consider:
1. How would you solve \( x^x = 4 \)?
2. What is the value of \( x \) if \( \log_2 8 = x^x \)?
3. Can you solve \( 3^x = x^3 \) by inspection?
4. How do you graph the function \( y = x^x \) and find where it intersects with \( y = 2 \)?
5. What are the properties of the function \( f(x) = x^x \) for \( x > 0 \)?

**Tip:** For equations of the form \( x^x = a \), sometimes a good starting point is to try simple numbers or consider if the base is a known exponent.

Explore how to solve the equation log4 16 = x^x step-by-step. This problem involves logarithms and exponential equations, demonstrating methods to find the solution. Suitable for advanced high school students and beyond.

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