University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.6 - Inverse Functions and Logarithms - Exercises - Page 50: 82


After $15$ years, the investment will double in value.

Work Step by Step

Interest is earned at the rate of $4.75\%$ compounded annually means that every year, the investment comes out as $104.75\%$ of the amount of the previous year. For example, here we have $\$500$ initially. After the first year, the investment values $104.75\%$ of $\$500$; in other words, its value now is $\$500\times1.0475$. After the second year, it values $104.75\%$ the investment of the previous year, which means its value now is $1.0475\times\$500\times1.0475=\$500\times(1.0475)^2$ Therefore, if we continue like that and call the value of the investment after $t$ years $y$, we can come up with the model to calculate that amount: $$y=500\times1.0475^t$$ For an investment of $\$500$ to double in value, meaning to find $t$ so that $y=1000$: $$500\times1.0475^t=1000$$ $$1.0475^t=2$$ - Take the $\log_{1.0475}$ of both sides: $$t=\log_{1.0475}2\approx14.936\approx15(years)$$
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