Intermediate Algebra (6th Edition)

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

Chapter 7 - Section 7.2 - Rational Exponents - Exercise Set - Page 425: 116

Answer

$393.4 \text{ million}$

Work Step by Step

Since $2015$ is $20$ years after $1995,$ then $x=20.$ Substituting $x=20$ in the given function, $ f(x)=25x^{23/25} ,$ then \begin{array}{l}\require{cancel} f(x)=25(20)^{23/25} .\end{array} Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} f(x)=25\sqrt[25]{20^{23}} .\end{array} Using the laws of exponents and the properties of radicals, the expression above is equivalent to \begin{array}{l}\require{cancel} f(x)=25\sqrt[25]{(2^2\cdot5)^{23}} \\ f(x)=25\sqrt[25]{2^{2(23)}\cdot5^{23}} \\ f(x)=25\sqrt[25]{2^{46}\cdot5^{23}} \\ f(x)=25\sqrt[25]{2^{25}\cdot2^{21}\cdot5^{23}} \\ f(x)=25(2)\sqrt[25]{2^{21}\cdot5^{23}} \\ f(x)=50\sqrt[25]{2^{21}\cdot5^{23}} \\ f(x)=50\sqrt[25]{2^{21}\cdot5^{21}\cdot5^2} \\ f(x)=50\sqrt[25]{(2\cdot5)^{21}\cdot5^2} \\ f(x)=50\sqrt[25]{10^{21}\cdot25} \\ f(x)\approx393.4 .\end{array} Hence, the number of cellular subscriptions, $f(x),$ is $ 393.4 \text{ million} .$
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