Answer
$x=2 \lor x=3$
Work Step by Step
Given: $2^{2x}-12.2^x+32=0\\(2^x-8)(2^x-4)=0$
We have $2^x-8=0\\2^x=8\\x=3$
and $2^x-4=0\\2^x=4\\x=2$
Check: $2^4-12.2^2+32=0\\16-12.4+32=0\\0=0$
$2^6-12.2^3+32=0\\64-12.8+32=0\\0=0$
Algebra 2 (1st Edition)
by Larson, Ron; Boswell, Laurie; Kanold, Timothy D.; Stiff, Lee
Published by
McDougal Littell
ISBN 10:
0618595414
ISBN 13:
978-0-61859-541-9
Chapter 7 Exponential and Logarithmic Functions - 7.6 Solve Exponential and Logarithmic Equations - 7.6 Exercises - Skill Practice - Page 520: 52
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