Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 6 - Radical Functions and Rational Exponents - 6-4 Rational Exponents - Practice and Problem-Solving Exercises - Page 386: 52

Answer

$8$

Work Step by Step

Note that: $4^{1.5}=4^{\frac{3}{2}}$ Since $2^2=4$, then: $4^{\frac{3}{2}}=(2^2)^{\frac{3}{2}}$ Now, recall the basic exponent property (pg. 360): $(a^m)^n=a^{mn}$ Applying this property, we get: \begin{align*} (2^2)^{\frac{3}{2}}&=2^{\frac{2\cdot 3}{2}}\\ &=2^{\frac{6}{2}}\\ &=2^{3}\\ &=8\end{align*}
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