Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.1 - Exponents and Scientific Notation - Exercise Set: 147



Work Step by Step

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.
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