To determine which of the two expressions, \(2^{(2^n)}\) or \(2^{(n^2)}\), is bigger, we can compare their growth rates as \(n\) increases:

1. **\(2^{(2^n)}\)** grows **exponentially** in \(n\). The exponent \(2^n\) itself grows extremely fast, doubling with each increase in \(n\).

2. **\(2^{(n^2)}\)** grows at a **quadratic exponential rate**, meaning the exponent \(n^2\) increases quadratically as \(n\) increases.

### Which grows faster?
- \(2^n\) grows much faster than \(n^2\) for large \(n\). Therefore, the exponent in \(2^{(2^n)}\) grows faster than that in \(2^{(n^2)}\), making \(2^{(2^n)}\) grow much faster overall.

Thus, **\(2^{(2^n)}\) is larger than \(2^{(n^2)}\)** for sufficiently large \(n\).

Would you like a detailed breakdown of this comparison for different values of \(n\), or further clarification?

---

Here are five follow-up questions to deepen your understanding:
1. How does \(2^{(2^n)}\) behave for small values of \(n\)?
2. Can you visualize the growth of \(2^n\) compared to \(n^2\)?
3. What happens to the expressions if \(n\) is negative?
4. Are there any specific real-world examples where such fast-growing functions are relevant?
5. How would the comparison change if logarithms were involved in the exponents?

**Tip:** Always compare the growth rates of the exponents to determine the relative size of exponential functions!

Discover which expression, 2^(2^n) or 2^(n^2), grows faster as n increases. This comparison delves into exponential growth rates and highlights how 2^(2^n) outpaces 2^(n^2) for large values of n. Suitable for high school students studying exponential functions.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation