Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 7 Exponential and Logarithmic Functions - Chapter Review - Page 542: 36



Work Step by Step

Let $y=ab^x$. Then our equations are: $2=ab^{-2}$ and $0.25=ab^{1}$. If we divide the second equation by the first one we get: $0.125=b^3$, thus $b=0.5$. Then $0.25=a(0.5)^1=0.5a\\a=0.5$. Thus $y=0.5\cdot(0.5)^x$
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