Algebra 2 (1st Edition)

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

Chapter 7 Exponential and Logarithmic Functions - 7.7 Write and Apply Exponential and Power Functions - Guided Practice for Examples 1, 2, and 3 - Page 531: 1

Answer

$y=3\cdot2^x$

Work Step by Step

Let $y=ab^x$. Then our equations are: $6=ab^1$ and $24=ab^3$. If we divide the second equation by the first one we get: $4=b^2$. Thus, because $b$ is positive: $b=2$. Then $6=a\cdot2^1=2a\\a=3$. Thus $y=3\cdot2^x$
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