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: 67

Answer

$x=\dfrac{13}{12}$

Work Step by Step

$\log_{5}(x+1)-\log_{5}(x-1)=2$ Combine the logarithms on the left side of the equation as a division: $\log_{5}\Big(\dfrac{x+1}{x-1}\Big)=2$ Write this equation in exponential form: $\Big(\dfrac{x+1}{x-1}\Big)=5^{2}$ $\Big(\dfrac{x+1}{x-1}\Big)=25$ Solve for $x$: $x+1=25(x-1)$ $x+1=25x-25$ $x-25x=-25-1$ $-24x=-26$ $x=\dfrac{-26}{-24}=\dfrac{13}{12}$
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