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 $x=-1$
College Algebra 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
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 - Page 404: 45
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