Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 9 - Section 9.6 - Exponential and Logarithmic Equations; Further Applications - 9.6 Exercises - Page 630: 44

Answer

$t=4$

Work Step by Step

Using the properties of logarithms, the given equation, $ \log_2(t+5)-\log_2 (t-1)=\log_2 3 $, is equivalent to \begin{align*}\require{cancel} \log_2\dfrac{t+5}{t-1}&=\log_2 3 &(\text{use }\log_b \dfrac{x}{y}=\log_b x-\log_b y) .\end{align*} Since $\log_b m=\log_b n $ implies $m=n$, the equation above implies \begin{align*}\require{cancel} \dfrac{t+5}{t-1}&=3 .\end{align*} Using the properties of equality, the equation above is equivalent to \begin{align*}\require{cancel} (\cancel{t-1})\cdot\dfrac{t+5}{\cancel{t-1}}&=3\cdot(t-1) \\\\ t+5&=3t-3 \\ 5+3&=3t-t \\ 8&=2t \\\\ \dfrac{8}{2}&=\dfrac{\cancel2t}{\cancel2} \\\\ 4&=t \\ t&=4 .\end{align*} Hence, the solution to the equation $ \log_2(t+5)-\log_2 (t-1)=\log_2 3 $ is $ t=4 $.
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