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 Example 4 - Page 531: 6

Answer

$y=\frac{3^{\ln_2{3.75}}}{4}\cdot x^{\ln_2{3.75}}$

Work Step by Step

Let $y=ax^b$. Then our equations are: $4=a3^b$ and $15=a6^b$. If we divide the second equation by the first one we get: $3.75=2^b$. Then: $b=\ln_2{3.75}$. Then $4=a3^{\ln_2{3.75}}\\a=\frac{3^{\ln_2{3.75}}}{4}$ Thus $y=\frac{3^{\ln_2{3.75}}}{4}\cdot x^{\ln_2{3.75}}$
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