Intermediate Algebra (6th Edition)

$7^{-11}$
We are given the numbers $7^{-11}$ and $7^{-13}$. We know that $a^{-n}=\frac{1}{a^{n}}$ (where a is a nonzero real number and n is a positive integer). Without calculating completely, we know that $7^{-13}=\frac{1}{7^{13}}=\frac{1}{7^{11}\times7^{2}}$. Therefore, both numbers will have the same numerator, but $7^{-13}$ will have a denominator that is $7^{2}=7\times7=49$ times greater than the denominator of $7^{-11}$. So, we know that $7^{-11}$ will be a larger number.