College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 4, Exponential and Logarithmic Functions - Chapter 4 Review - Concept Check - Page 425: 73


$x=\frac{\log 3+\log 5}{4\log 3-2\log 5}\approx 2.303600$

Work Step by Step

We solve for $x$: $5^{2x+1} =3^{4x-1}$ $(2x+1)\log 5=(4x-1)\log 3$ $2x\log 5+\log 5=4x\log 3-\log 3$ $2x\log 5-4x\log3=-\log 3-\log 5$ $x(2 \log 5-4\log 3)=-\log 3-\log 5$ $x=\frac{\log 3+\log 5}{4\log 3-2\log 5}\approx 2.303600$
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