To determine if \( n^{\log 5} \) is \( \Theta(n^3) \), we can compare the growth rates of the two expressions using logarithms.

### Step-by-step approach:

1. **Expression 1**: \( n^{\log 5} \)

   This is an exponential function where the base is \( n \) and the exponent is \( \log 5 \). Since \( \log 5 \) is a constant, let's approximate its value:

   \[
   \log 5 \approx 0.69897
   \]

   So \( n^{\log 5} \) grows at a rate of approximately \( n^{0.69897} \).

2. **Expression 2**: \( n^3 \)

   This grows at a rate of \( n^3 \), which is clearly much faster than \( n^{0.69897} \) as \( n \) increases.

### Comparison of growth rates:

- \( n^{\log 5} \approx n^{0.69897} \)
- \( n^3 \)

Clearly, \( n^{0.69897} \) grows much slower than \( n^3 \).

### Conclusion:

Since \( n^{\log 5} \) grows slower than \( n^3 \) as \( n \to \infty \), \( n^{\log 5} \) is **not** \( \Theta(n^3) \). In fact, \( n^{\log 5} = o(n^3) \), meaning it grows strictly slower than \( n^3 \).

### Final answer: 

No, \( n^{\log 5} \) is not \( \Theta(n^3) \).

Would you like more details or have any questions?

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Here are 5 related questions you might find helpful:
1. How do you formally prove two functions are in \( \Theta \)-notation?
2. What does it mean for a function to be in \( o(g(n)) \)?
3. How can we compare growth rates using Big-O, Big-\( \Omega \), and Big-\( \Theta \) notations?
4. What are typical methods for determining asymptotic bounds?
5. How does logarithmic growth compare to polynomial growth in terms of complexity?

**Tip**: To determine the asymptotic behavior of functions, always check how they compare as \( n \to \infty \) by analyzing limits or logarithmic comparisons.

Explore the comparison between n^log5 and n^3 in terms of asymptotic growth. This problem analyzes whether n^log5 is Θ(n^3), using logarithmic approximations and growth rate comparisons, and concludes that n^log5 grows slower than n^3.

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