Notes
Compound Interest Formula
\(A=P\left(1+\frac{r}{n} \right)^{nt}\)
\(A=\)Amount at time t
\(P=\)Principle (amount at time t=0)
\(n=\)compoundings per year
\(r=\)annual interest rate (\(3\%=.03\) for example)
\(t=\)time in years
Values of \(n\)
Monthly \( \rightarrow n=12\)
Quarterly \( \rightarrow n=4\)
Daily \( \rightarrow n=365\)
Weekly \( \rightarrow n=52\)
Yearly \( \rightarrow n=1\)
Annually \( \rightarrow n=1\)
Semi-annually \( \rightarrow n=2\)
Continuously Compounded Interest Formula
\(A=P e^{rt}\)
Interest Earned
\(I=A-P\)
Problems & Videos
\(\textbf{1)}\) You place $4,000 in a bank account with 2.5% interest rate compounded monthly. How much will you have in the account after 4 years?
The answer is \( $4{,}420.22 \)
\(\,\,\,\,\,\,A=P\left(1+\frac{r}{n}\right)^{nt}\)
\(\,\,\,\,\,\,A=4{,}000\left(1+\frac{.025}{12}\right)^{12 \cdot 4}\)
\(\,\,\,\,\,\,A=4{,}000\left(1+0.00208333\right)^{48}\)
\(\,\,\,\,\,\,A=4{,}000\left(1.00208333\right)^{48}\)
\(\,\,\,\,\,\,A=4{,}000\cdot 1.105056\)
\(\,\,\,\,\,\,\)The answer is \( $4{,}420.22 \)
\(\textbf{2)}\) How much money do you have after 5 years if you invest $6,000 at 3.5% interest compounded quarterly?
The answer is \( $7{,}142.04 \)
\(\,\,\,\,\,\,A=P\left(1+\frac{r}{n}\right)^{nt}\)
\(\,\,\,\,\,\,A=6{,}000\left(1+\frac{.035}{4}\right)^{(4)(5)}\)
\(\,\,\,\,\,\,A=6{,}000\left(1.00875\right)^{20}\)
\(\,\,\,\,\,\,A=6{,}000 \cdot 1.19034\)
\(\,\,\,\,\,\,A=6{,}000 \cdot 1.19034\)
\(\,\,\,\,\,\,\)The answer is \( $7{,}142.04 \)
\(\textbf{3)}\) Alex puts $800.00 into an account. The account earns 4.1% interest, compounded monthly. How much will be in the account after 7 years?
The answer is \( $1{,}065.42 \)
\(\,\,\,\,\,\,A=P\left(1+\frac{r}{n}\right)^{nt}\)
\(\,\,\,\,\,\,A=800\left(1+\frac{.041}{12}\right)^{12 \cdot 7}\)
\(\,\,\,\,\,\,A=800\left(1+0.00341667\right)^{84}\)
\(\,\,\,\,\,\,A=800\left(1.00341667\right)^{84}\)
\(\,\,\,\,\,\,A=800\cdot 1.331773\)
\(\,\,\,\,\,\,\)The answer is \( $1{,}065.42 \)
\(\textbf{4)}\) Stevie puts $800.00 into an account. The account earns 4.1% interest compounded continously. How much will be in the account after 7 years?
The answer is \( $1{,}065.94 \)
\(\,\,\,\,\,\,A=Pe^{rt}\)
\(\,\,\,\,\,\,A=800e^{(0.041)(7)}\)
\(\,\,\,\,\,\,A=800e^{0.287}\)
\(\,\,\,\,\,\,A=800(1.332424)\)
\(\,\,\,\,\,\,A=1065.94\)
\(\,\,\,\,\,\,\)The answer is \( $1{,}065.94 \)
\(\textbf{5)}\) You need $30,000 in 5 years. How much money should you place in an account with 3% interest compounded continuously?
The answer is \( $25{,}821.24 \)
\(\,\,\,\,\,\,A=Pe^{rt}\)
\(\,\,\,\,\,\,30{,}000=Pe^{(.03)(5)}\)
\(\,\,\,\,\,\,30{,}000=Pe^{.15}\)
\(\,\,\,\,\,\,30{,}000=Pe^{.15}\)
\(\,\,\,\,\,\,30{,}000=P(1.161834)\)
\(\,\,\,\,\,\,\displaystyle\frac{30{,}000}{1.161834}=P\)
\(\,\,\,\,\,\,25821.24468=P\)
\(\,\,\,\,\,\,\)The answer is \( $25{,}821.24 \)
\(\textbf{6)}\) How long would it take $3,000 to triple if you earned 4% interest compounded continuously?
The answer is \( 27.47\) years
\(\,\,\,\,\,\,A=Pe^{rt}\)
\(\,\,\,\,\,\,9{,}000=3{,}000e^{.04t}\)
\(\,\,\,\,\,\,3=e^{.04t}\)
\(\,\,\,\,\,\,\ln{3}=\ln{e^{.04t}}\)
\(\,\,\,\,\,\,\ln{3}=.04t\)
\(\,\,\,\,\,\,\displaystyle\frac{\ln{3}}{.04}=t\)
\(\,\,\,\,\,\,27.46530=t\)
\(\,\,\,\,\,\,\)The answer is \( 27.47\) years
\(\textbf{7)}\) Kay invests $8,294 in an account that pays an annual interest rate of 4% compounded monthly. What will the account balance be after 6 years?
The answer is \( $10{,}539.53 \)
\(\,\,\,\,\,\,A=P\left(1+\frac{r}{n}\right)^{nt}\)
\(\,\,\,\,\,\,A=8{,}294\left(1+\frac{.04}{12}\right)^{\left(12\right)\left(6\right)}\)
\(\textbf{8)}\) Ian invests $3,925 in an account with a fixed annual interest rate of 6.5% compounded twice per year. What will the account balance be after 10 years?
The answer is \( $7{,}441.16 \)
\(\,\,\,\,\,\,A=P\left(1+\frac{r}{n}\right)^{nt}\)
\(\,\,\,\,\,\,A=3925\left(1+\frac{.065}{2}\right)^{\left(2\right)\left(10\right)}\)
\(\textbf{9)}\) Steph invests $7,498 in an account with an annual interest rate of 4.2% compounded continuously. What will the account balance be after 16 years?
The answer is \( $14{,}682.21 \)
\(\,\,\,\,\,\,A=Pe^{rt}\)
\(\,\,\,\,\,\,A=7498e^{.042\cdot 16}\)
\(\textbf{10)}\) Kelsey invests $3,225 in a savings account with an annual interest rate of 7% compounded continuously. What will the account balance be after 12 years?
The answer is \( $7{,}470.28 \)
\(\,\,\,\,\,\,A=Pe^{rt}\)
\(\,\,\,\,\,\,A=3{,}225\:\:e^{.07\cdot 12}\)
\(\textbf{11)}\) Ashton invests a sum of money in an account with an annual interest rate of 6.32% compounded monthly. After 6 years, the balance reaches $28,449.75. What was the amount of the initial investment?
The answer is \( $19{,}490.64 \)
\(\,\,\,\,\,\,A=P\left(1+\frac{r}{n}\right)^{nt}\)
\(\,\,\,\,\,\,28449.75=P\left(1+\frac{0.0632}{12}\right)^{\left(12\right)\left(6\right)}\)
\(\,\,\,\,\,\,P=\displaystyle \frac{28449.75}{\left(1+\frac{0.0632}{12}\right)^{\left(12\right)\left(6\right)}}\)
\(\textbf{12)}\) Izzy invests a sum of money in an account with an annual interest rate of 3.51% compounded weekly. After 8 years, the balance reaches $10,376.23. What was the amount of the initial investment?
The answer is \( $7{,}836.66 \)
\(\,\,\,\,\,\,A=P\left(1+\frac{r}{n}\right)^{nt}\)
\(\,\,\,\,\,\,10376.24=P\left(1+\frac{.0351}{52}\right)^{\left(52\right)\left(8\right)}\)
\(\,\,\,\,\,\,P=\displaystyle \frac{10376.24}{\left(1+\frac{.0351}{52}\right)^{\left(52\right)\left(8\right)}}\)
\(\textbf{13)}\) Jen invests a sum of money in an account with a fixed annual interest rate of 3.45% compounded continuously. After 16 years, the balance reaches $4,567.89. What was the amount of the initial investment?
The answer is \( $2{,}630.18 \)
\(\,\,\,\,\,\,A=Pe^{rt}\)
\(\,\,\,\,\,\,4567.89=Pe^{.0345\cdot 16}\)
\(\,\,\,\,\,\,P=\displaystyle \frac{4567.89}{e^{.0345\cdot 16}}\)
\(\textbf{14)}\) Jason invests a sum of money in an account with an annual interest rate of 7.28% compounded continuously. After 11 years, the balance reaches $18,153.16. What was the amount of the initial investment?
The answer is \( $8{,}150.22 \)
\(\,\,\,\,\,\,A=Pe^{rt}\)
\(\,\,\,\,\,\,18153.16=Pe^{.0728\cdot 11}\)
\(\,\,\,\,\,\,P=\displaystyle \frac{18153.16}{e^{.0728\cdot 11}}\)
\(\textbf{15)}\) David invests $4,495 in an account with a fixed annual interest rate compounded continuously. After 15 years, the balance reaches $9,515.92. What is the interest rate of the account?
The answer is \( 5\% \)
\(\,\,\,\,\,\,A=Pe^{rt}\)
\(\,\,\,\,\,\,9515.92=4495e^{r\cdot 15}\)
\(\,\,\,\,\,\,e^{15r}=\frac{9515.92}{4495}\)
\(\,\,\,\,\,\,15r=\ln{\frac{9515.92}{4495}}\)
\(\,\,\,\,\,\,r=\displaystyle \frac{\ln{\frac{9515.92}{4495}}}{15}\)
\(\,\,\,\,\,\,r=.05\)
\(\textbf{16)}\) Olivia invests $2,854 in an account with an annual interest rate of 6.5% compounded continuously. How long will it take for the account balance to reach $5,122.90?
The answer is \( 9 \) years
\(\,\,\,\,\,\,A=Pe^{rt}\)
\(\,\,\,\,\,\,5122.90=2854e^{.065\cdot t}\)
\(\,\,\,\,\,\,e^{.065 t}=\frac{5122.9}{2854}\)
\(\,\,\,\,\,\,.065 t=\ln{\frac{5122.9}{2854}}\)
\(\,\,\,\,\,\, t=\displaystyle \frac{\ln{\frac{5122.9}{2854}}}{0.065}\)
\(\,\,\,\,\,\, t=9\)
\(\textbf{17)}\) Mike invested P dollars. The value of his invested decreased by 5.5% each year. At the end of the first year, the value of Mike’s investment was $5670. What was P?
The answer is \( $6000 \)
\(\textbf{18)}\) What if your ancestor put $1 in the bank in the year 0 at 2% interest compounded monthly. What would it be worth today?
The answer is approximately \( $346,394,275,535,600,000. \)
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