Respuesta :
Recall the formula for compound interest
[tex]\begin{gathered} A=P\Big(1+\frac{r}{n}\Big)^{nt} \\ \text{where} \\ A\text{ is the accrued amount} \\ P\text{ is the principal amount} \\ r\text{ is the rate as a decimal} \\ n\text{ is the number of compounding period per year} \\ t\text{ is the time in years} \end{gathered}[/tex]Using the formula, rearrange the equation and solve in terms of time t
[tex]\begin{gathered} A=P\Big{(}1+\frac{r}{n}\Big{)}^{nt} \\ \frac{A}{P}=\frac{\cancel{P}\Big{(}1+\frac{r}{n}\Big{)}^{nt}}{\cancel{P}} \\ \frac{A}{P}=\Big{(}1+\frac{r}{n}\Big{)}^{nt} \\ \ln \Big(\frac{A}{P}\Big)=\ln \Big(1+\frac{r}{n}\Big)^{nt} \\ \frac{\ln \Big{(}\frac{A}{P}\Big{)}}{n\ln \Big{(}1+\frac{r}{n}\Big{)}}=\frac{nt\ln \Big{(}1+\frac{r}{n}\Big{)}}{n\ln \Big{(}1+\frac{r}{n}\Big{)}} \\ t=\frac{\ln \Big{(}\frac{A}{P}\Big{)}}{n\ln \Big{(}1+\frac{r}{n}\Big{)}} \end{gathered}[/tex]Substitute the given values and we get
A = $2700, P = $1100, n = 12 (12 months in a year), r = 0.045 (from 4.5%)
[tex]\begin{gathered} t=\frac{\ln\Big{(}\frac{A}{P}\Big{)}}{n\ln\Big{(}1+\frac{r}{n}\Big{)}} \\ t=\frac{\ln \Big{(}\frac{2700}{1100}\Big{)}}{(12)\ln \Big{(}1+\frac{0.045}{12}\Big{)}} \\ t=19.99 \end{gathered}[/tex]Rounding the answer to the nearest tenth, the time it takes is 20 years.
