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: 54

Answer

$x=3$

Work Step by Step

$\log_{4}(x+2)+\log_{4}3=\log_{4}5+\log_{4}(2x-3)$ Combine the logarithms on the left and right side of the equation as products: $\log_{4}(3x+6)=\log_{4}(10x-15)$ Since $\log$ is one to one, the equation becomes: $3x+6=10x-15$ Solve for $x$: $3x-10x=-15-6$ $-7x=-21$ $x=3$
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