Answer
After $\approx2.4$ years.
Work Step by Step
From part a): $y=200(0.75)^t$. Hence if $y=100$, then the equation is: $100=200(0.75)^t\\0.5=(0.75)^t\\t=\log_{0.75}{0.5}\approx2.4$
Algebra 2 (1st Edition)
by Larson, Ron; Boswell, Laurie; Kanold, Timothy D.; Stiff, Lee
Published by
McDougal Littell
ISBN 10:
0618595414
ISBN 13:
978-0-61859-541-9
Chapter 7 Exponential and Logarithmic Functions - 7.2 Graph Exponential Decay Functions - 7.2 Exercises - Problem Solving - Page 490: 31c
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