Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Practice Exercises - Page 440: 79

Answer

$$ y = \frac{{\ln 2}}{{\ln \left( {3/2} \right)}}$$

Work Step by Step

$$\eqalign{ & {3^y} = {2^{y + 1}} \cr & {\text{take the natural logarithm on both sides}} \cr & \ln {3^y} = \ln {2^{y + 1}} \cr & {\text{use the power property}} \cr & y\ln 3 = \left( {y + 1} \right)\ln 2 \cr & y\ln 3 = y\ln 2 + \ln 2 \cr & {\text{solve for }}y \cr & y\ln 3 - y\ln 2 = \ln 2 \cr & y\left( {\ln 3 - \ln 2} \right) = \ln 2 \cr & y = \frac{{\ln 2}}{{\ln 3 - \ln 2}} \cr & y = \frac{{\ln 2}}{{\ln \left( {3/2} \right)}} \cr} $$
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