Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.4 - Exponential Change and Separable Differential Equations - Exercises 7.4 - Page 401: 30

Answer

$1250$

Work Step by Step

Consider the exponential growth as: $y=y_0e^{kt}$ As we are given that $y(3)=10,000$ and $y(5)=40,000$ Then, we have $10,000=y_0 e^{3k}$ and $40,000=y_0 e^{5k}$ From the above two equations, we get $y_0 e^{5k}=4y_0 e^{3k}$ or, $e^{2k}=4$ or, $k= \ln 2$ $y=y_0e^{(\ln 2)t} \implies 10,000=(y_0) e^{3 \ln 2}$ or, $10,000=y_0e^{\ln (2^3)}$ so, $10,000=(8) y_0 \implies y_0=1250$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.