Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Review - Exercises - Page 389: 57

Answer

$x=5$

Work Step by Step

$3^{2x-7}=27$ Rewrite $27$ as $3^{3}$: $3^{2x-7}=3^{3}$ If $3^{2x-7}=3^{3}$, then $2x-7=3$ $2x-7=3$ Solve for $x$: $2x=3+7$ $2x=10$ $x=\dfrac{10}{2}$ $x=5$
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