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: 45

Answer

$x=\pm 1$

Work Step by Step

We need to solve: $x^{2}2^{x}-2^{x}=0$ We factor: $2^{x}(x^{2}-1)=0$ $2^x(x-1)(x+1)=0$ $2^{x}=0$ or $x-1=0$ or $x+1=0$ $x=\log_2 0$=no solution, or $x=1$, or $x=-1$
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