Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 9 - Section 9.2 - Exponential Functions - 9.2 Exercises - Page 597: 32



Work Step by Step

We are given the equation $9^{2x-8}=27^{x-4}$. First, we must write both sides using the same base. $((3)^{2})^{2x-8}=3^{4x-16}$ $((3)^{3})^{x-4}=3^{3x-12}$ So, $3^{4x-16}=3^{3x-12}$. Next, we must take the natural log of both sides. $ln(3^{4x-16})=ln(3^{3x-12})$ $(4x-16)ln(3)=(3x-12)ln(3)$ Divide both sides by $ln(3)$. $4x-16=3x-12$ Subtract $3x$ from both sides. $x-16=-12$ Add 16 to both sides. $x=4$
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