Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 4 - Section 4.5 - Exponential and Logarithmic Functions - 4.5 Exercises - Page 368: 5

Answer

$x=\dfrac{3}{2}$

Work Step by Step

$5^{2x-3}=1$ Rewrite this equation as $5^{2x}\cdot5^{-3}=1$ $5^{2x}\cdot5^{-3}=1$ Take $5^{-3}$ to divide the right side of the equation: $5^{2x}=\dfrac{1}{5^{-3}}$ $5^{2x}=5^{3}$ Use the one-to-one property to solve this equation: $2x=3$ Solve for $x$: $x=\dfrac{3}{2}$
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