College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 4, Exponential and Logarithmic Functions - Section 4.5 - Exponential and Logarithmic Functions - 4.5 Exercises: 43

Answer

$x=1$

Work Step by Step

We need to solve: $2^{x}-10(2^{-x})+3=0$ We multiply through by $2^x$: $2^x(2^{x}-10(2^{-x})+3=0)$ $2^{2x}-10+3* 2^{x}=0$ $(2^{x}+5)(2^{x}-2)=0$ $(2^{x}+5)=0$ or $(2^{x}-2)=0$ $2^x=-5$ or $2^x=2$ $x=\log_2 -5$=no solution or $x=\log_2 2=1$
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