1. The problem asks us to compare two expressions: $N^{1.1}$ and $100^N$. 2. We want to determine which is larger for a given value of $N$. 3. Recall the properties of exponents: $a^b$ grows faster as $b$ increases if $a>1$. 4. Here, $N^{1.1}$ means $N$ raised to the power of $1.1$, which grows polynomially. 5. $100^N$ means $100$ raised to the power of $N$, which grows exponentially. 6. Exponential growth ($100^N$) eventually surpasses any polynomial growth ($N^{1.1}$) as $N$ becomes large. 7. For small values of $N$, $N^{1.1}$ might be larger, but as $N$ increases, $100^N$ becomes much larger. 8. To see this, consider $N=1$: $1^{1.1}=1$ and $100^1=100$, so $100^N$ is larger. 9. For $N=0.1$: $0.1^{1.1}$ is about $0.079$, and $100^{0.1}$ is about $1.58$, so $100^N$ is still larger. 10. Therefore, $100^N$ is larger than $N^{1.1}$ for all positive $N$ except possibly very close to zero. Final answer: $100^N$ is larger than $N^{1.1}$ for all positive $N$ except possibly near zero.