Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 4 - Section 4.5 - Exponential and Logarithmic Functions - 4.5 Exercises - Page 369: 90

Answer

a. $\$ 7328.73$ b. about $3\displaystyle \frac{1}{2}$ years

Work Step by Step

If a principal $P$ is invested in an account paying an annual interest rate $r$, and the interest is compounded continuously, then the amount after t years is $A(t)=Pe^{rt}$ -------------- a. $A(2)=6500e^{0.06(2)}\approx\$ 7328.73$ b. Solve for t after inserting given values $8000=6500e^{0.06t} \qquad.../\div 6500$ $\displaystyle \frac{16}{13}=e^{0.06t} \qquad$ ... apply ln() to both sides $\displaystyle \ln(\frac{16}{13})=0.06t \qquad$ ... $/\div 0.06$ $t=\displaystyle \frac{\ln(\frac{16}{13})}{0.06}\approx 3.46$. The investment grows to $\$ 8000$ in about $3\displaystyle \frac{1}{2}$ years.
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.