## University Calculus: Early Transcendentals (3rd Edition)

After $15$ years, the investment will double in value.
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: $$t=\log_{1.0475}2\approx14.936\approx15(years)$$