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: 20



Work Step by Step

We are given the exponential equation $8^{1-x}=4^{x+2}$. We can express each side using a common base and then solve for $x$. $8^{1-x}=(2^{3})^{1-x}=2^{3-3x}$ $4^{x+2}=(2^{2})^{x+2}=2^{2x+4}$ $2^{3-3x}=2^{2x+4}$ Take the natural log of both sides. $ln(2^{3-3x})=ln(2^{2x+4})$ $(3-3x)ln(2)=(2x+4)ln(2)$ Divide both sides by $ln(2)$. $3-3x=2x+4$ Add $3x$ to both sides. $3=5x+4$ Subtract 4 from both sides. $5x=-1$ Divide both sides by 5. $x=-\frac{1}{5}$
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