Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 3 - Section 3.4 - Exponential and Logarithmic Equations - Exercise Set - Page 489: 105

Answer

There is $ 1\%$ of surface sunlight at $118$ feet. The point $(118,1)$ is on the graph.

Work Step by Step

We want to find the x for which $ f(x)=1.$ $ 1=20(0.975)^{x}\quad $ ... $/\div 20$ $\displaystyle \frac{1}{20}=0.975^{x}\quad $ ... $/\ln(...)$ $\displaystyle \ln\frac{1}{20}=x\ln 0.975\quad $ ... $/\div\ln 0.975$ $ x=\displaystyle \frac{\ln(1/20)}{\ln 0.975}\approx 118.325$ There is $ 1\%$ of surface sunlight at $118$ feet. The point $(118,1)$ is on the graph.
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