## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter R - Algebra Reference - R.6 Exponents - R.6 Exercises - Page R-25: 16

#### Answer

$\displaystyle \frac{1}{3z^{7}}$

#### Work Step by Step

In each step, we use one of the Properties of Exponents... P0. $a^{0}=1, a^{1}=a$ P1. $a^{m}\cdot a^{n}=a^{m+n}$ P2. $\displaystyle \frac{a^{m}}{a^{n}}=a^{m-n}$ , $\displaystyle \frac{1}{a^{n}}=a^{-n}$ P3. $(a^{m})^{n}=a^{mn}$ P4. $(ab)^{m}=a^{m}b^{m}$ P5. $(\displaystyle \frac{a}{b})^{m}=\frac{a^{m}}{b^{m}}$ P6. Rational exponents: $a^{m/n}=(a^{1/n})^{m}$=$\sqrt[n]{a^{m}}$ ----------------------------- $\displaystyle \frac{(3z^{2})^{-1}}{z^{5}}$= ... P$4$... $=\displaystyle \frac{3^{-1}(z^{2})^{-1}}{z^{5}}$= ... P$3$... =$\displaystyle \frac{3^{-1}z^{-2}}{z^{5}}=3^{-1}\cdot\frac{z^{-2}}{z^{5}}$= ... P$2$... $=3^{-1}z^{-2-5}=3^{-1}z^{-7}$= ... P$2$... $=\displaystyle \frac{1}{3}\cdot\frac{1}{z^{7}}$ $= \displaystyle \frac{1}{3z^{7}}$

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