College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter P, Prerequisites - Section P.3 - Integer Exponents and Scientific Notation - P.3 Exercises: 30

Answer

(a) $\dfrac{x^6}{y^{12}}$ (b) $\dfrac{x^9}{8y^{14}}$ (c) $\dfrac{1}{32ab^8}$

Work Step by Step

RECALL: (1) $(ab)^m=a^mb^m$ (2) $(a^m)^n=a^{mn}$ (3) $a^{-m} = \dfrac{1}{a^m}$ (4) $\dfrac{a^m}{a^n} = a^{m-n}, a\ne0$ (5) $a^m\cdot a^n=a^{m+n}$ (a) Use rule (1) above to obtain: $=(x^{-2})^{-3}(y^4)^{-3}$ Use rule (2) above to obtain: $=x^{-2(-3)}y^{4(-3)} \\=x^{6}y^{-12}$ Use rule (3) above to obtain: $=x^{6} \cdot \dfrac{1}{y^{12}} \\=\dfrac{x^6}{y^{12}}$ (b) use rule (1) above to obtain: $=(y^2)^{-1}(2^{-3})(x^{-3})^{-3}(y^4)^{-3}$ Use rule (2) above to obtain: $=y^{2(-1)}(2^{-3})(x^{-3(-3)})y^{4(-3)} \\=y^{-2}(2^{-3})x^{9}y^{-12}$ Use rule (5) above to obtain: $\\=y^{-2}(2^{-3})x^{9}y^{-12} \\=(2^{-3})x^9y^{-2+(-12)} \\=(2^{-3})x^9y^{-14}$ Use rule (3) above to obtain: $=\dfrac{1}{2^3} \cdot x^9 \cdot \dfrac{1}{y^{14}} \\=\dfrac{1}{8} \cdot x^9 \cdot \dfrac{1}{y^{14}} \\=\dfrac{x^9}{8y^{14}}$ (c) Use rule (1) above to obtain: $=\left(\dfrac{2^{-3} (a^{-1})^{-3}}{(b^{-2})^{-3}}\right) \left(\dfrac{(b^{-1})^2}{2^2(a^2)^2}\right)$ Use rule (2) above to obtain: $=\left(\dfrac{2^{-3} (a^{-1(-3)}}{(b^{-2(-3)})}\right) \left(\dfrac{(b^{-1(2)})}{4(a^{2(2)}}\right) \\=\left(\dfrac{2^{-3} a^{3}}{b^{6}}\right) \left(\dfrac{b^{-2}}{4a^{4}}\right) \\=\dfrac{2^{-3}a^{3}b^{-2}}{4a^4b^6}$ Use rule (4) above to obtain: $=\dfrac{2^{-3}}{4}a^{3-4}b^{-2-6} \\=\dfrac{2^{-3}}{4}a^{-1}b^{-8}$ Use rule (3) above to obtain: $=\dfrac{1}{2^3(4)} \cdot \dfrac{1}{a} \cdot \dfrac{1}{b^{8}} \\=\dfrac{1}{8(4)ab^8} \\=\dfrac{1}{32ab^8}$
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