Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.5 Exercises - Page 362: 24


a. $x=-1$, $x=3$ b. $x=\frac{1}{3}$

Work Step by Step

a. $\log_{3}x+\log_{3}(x-2)=1$ $\log_{3}(x(x-2))=1$ $3^1=x^2-2x$ $3=x^2-2x$ $0=x^2-2x-3$ $0=(x+1)(x-3)$ $x=-1$, $x=3$ --------- b. $\log_{10}(x+3)-log_{10}x=1$ $\log_{10}(\frac{x+3}{x})=1$ $10^1=\frac{x+3}{x}$ $10x=x+3$ $9x=3$ $x=\frac{1}{3}$
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