College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 4 - Exponential and Logarithmic Functions - Exercise Set 4.4 - Page 490: 19



Work Step by Step

We are given the exponential equation $8^{x+3}=16^{x-1}$. We can express each side using a common base and then solve for $x$. $8^{x+3}=(2^{3})^{x+1}=2^{3x+3}$ $16^{x-1}=(2^{4})^{x-1}=2^{4x-4}$ $2^{3x+3}=2^{4x-4}$ Take the natural log of both sides. $ln(2^{3x+3})=ln(2^{4x-4})$ $(3x+3)ln(2)=(4x-4)ln(2)$ Divide both sides by $ln(2)$. $3x+3=4x-4$ Subtract $3x$ from both sides. $3=x-4$ Add 4 to both sides. $x=7$
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