Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 4 - Review - Exercises: 73

Answer

$x=-\dfrac{\log3+\log5}{2\log5-4\log3}\approx2.303599$

Work Step by Step

$5^{2x+1}=3^{4x-1}$ Apply $\log$ to both sides of the equation: $\log5^{2x+1}=\log3^{4x-1}$ Take the exponents down to multiply in front of their respective $\log$: $(2x+1)\log5=(4x-1)\log3$ Evaluate the products on both sides: $2x\log5+\log5=4x\log3-\log3$ Take $4x\log3$ to the left side and $\log5$ to the right side: $2x\log5-4x\log3=-\log3-\log5$ Take out common factor $x$ from the left side: $x(2\log5-4\log3)=-\log3-\log5$ Solve for $x$: $x=-\dfrac{\log3+\log5}{2\log5-4\log3}\approx2.303599$
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