To determine whether $n^{\log 5}$ is $\Theta(n^3)$, we need to compare the growth rates of these two functions.

Step-by-Step Comparison

1. Expression 1: $n^{\log 5}$
   The term $\log 5$ is the logarithm of 5 in base 10 (or base 2, depending on context, though base 2 is common in theoretical computer science). So, $n^{\log 5}$ means that the exponent of $n$ is approximately $\log 5$. Numerically, $\log 5 \approx 2.32$.

   Thus, $n^{\log 5} \approx n^{2.32}$.

2. Expression 2: $n^3$
   This is a simple polynomial function where $n$ is raised to the power of 3.

Comparison

• We are comparing $n^{2.32}$ and $n^3$. Clearly, $n^{2.32}$ grows slower than $n^3$, because $2.32 < 3$.

Conclusion

Since $n^{2.32}$ grows slower than $n^3$, the function $n^{\log 5}$ is not $\Theta(n^3)$. In fact, we can say:

• $n^{\log 5} = O(n^3)$ (since it grows slower than $n^3$),
• but $n^{\log 5} \neq \Omega(n^3)$ (since it does not grow as fast as $n^3$).

Therefore, $n^{\log 5}$ is not $\Theta(n^3)$.

Would you like further details on complexity comparisons?

Here are five related questions to explore further:

1. What is the meaning of $\Theta$, $O$, and $\Omega$ in asymptotic notation?
2. How can we compare the growth rates of logarithmic and polynomial functions?
3. What are the key differences between $O$-notation and $\Theta$-notation?
4. How does base conversion in logarithms affect asymptotic growth rates?
5. Can we have functions that are $O(n^2)$ but not $\Theta(n^2)$?

Tip: When comparing functions for asymptotic behavior, always check both the upper and lower bounds to establish whether they are truly $\Theta$.