College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Test - Page 472: 13



Work Step by Step

Using the laws of exponents, the given equation, $ 16^{2x+1}=8^{3x} ,$ is equivalent to \begin{array}{l}\require{cancel} (2^4)^{2x+1}=(2^3)^{3x} \\\\ 2^{4(2x+1)}=2^{3(3x)} .\end{array} Since the bases are the same, the exponents can be equated. Hence, the solution is \begin{array}{l}\require{cancel} 4(2x+1)=3(3x) \\\\ 8x+4=9x \\\\ 8x-9x=-4 \\\\ -x=-4 \\\\ x=4 .\end{array}
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