156k views
3 votes
Suppose you have $1100 deposited at 4.5% compounded monthly. About long will it take your balance toincrease to $2700? Round your answer to the nearest tenth of a year.years

1 Answer

5 votes

Recall the formula for compound interest


\begin{gathered} A=P\Big(1+(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}

Using the formula, rearrange the equation and solve in terms of time t


\begin{gathered} A=P\Big{(}1+(r)/(n)\Big{)}^(nt) \\ (A)/(P)=\frac{\cancel{P}\Big{(}1+(r)/(n)\Big{)}^(nt)}{\cancel{P}} \\ (A)/(P)=\Big{(}1+(r)/(n)\Big{)}^(nt) \\ \ln \Big((A)/(P)\Big)=\ln \Big(1+(r)/(n)\Big)^(nt) \\ \frac{\ln \Big{(}(A)/(P)\Big{)}}{n\ln \Big{(}1+(r)/(n)\Big{)}}=\frac{nt\ln \Big{(}1+(r)/(n)\Big{)}}{n\ln \Big{(}1+(r)/(n)\Big{)}} \\ t=\frac{\ln \Big{(}(A)/(P)\Big{)}}{n\ln \Big{(}1+(r)/(n)\Big{)}} \end{gathered}

Substitute the given values and we get

A = $2700, P = $1100, n = 12 (12 months in a year), r = 0.045 (from 4.5%)


\begin{gathered} t=\frac{\ln\Big{(}(A)/(P)\Big{)}}{n\ln\Big{(}1+(r)/(n)\Big{)}} \\ t=\frac{\ln \Big{(}(2700)/(1100)\Big{)}}{(12)\ln \Big{(}1+(0.045)/(12)\Big{)}} \\ t=19.99 \end{gathered}

Rounding the answer to the nearest tenth, the time it takes is 20 years.

User Asoub
by
7.2k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories