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Solved Examples: Continuous Compound Interest

Formulas Used: Mathematics of Finance
Verify Answers with Calculator: Financial Mathematics Calculators

NOTE: Unless instructed otherwise;
For all financial calculations, do not round until the final answer.
Do not round intermediate calculations. If it is too long, write it to "at least" 5 decimal places.
Round your final answer to $2$ decimal places.
Make sure you include your unit.

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For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

(1.) How long will it take money to double if it is invested at 9% interest rate compounded continuously?
Round to the nearest integer.

$P = ? \\[3ex] A = 2P...double \\[3ex] r = 9\% = \dfrac{9}{100} = 0.09 \\[5ex] t = ? \\[3ex] t = \dfrac{\ln \left(\dfrac{A}{P}\right)}{r} \\[7ex] t = \dfrac{\ln \left(\dfrac{2P}{P}\right)}{0.09} \\[7ex] = \dfrac{\ln 2}{0.09} \\[5ex] = \dfrac{0.693147181}{0.09} \\[5ex] = 7.70163534 \\[3ex] \approx 8\:years \\[3ex]$ Ceteris paribus, it will take about $8\:years$ for any sum of money invested at $9\%$ to double if the interest is compounded continuously.
(2.) If the population of SamDom For Peace City is growing at 1.5% compounded continuously, how long will it take the population to triple?

$P = ? \\[3ex] A = 3P...triple \\[3ex] r = 1.5\% = \dfrac{1.5}{100} = 0.015 \\[5ex] t = ? \\[3ex] t = \dfrac{\ln \left(\dfrac{A}{P}\right)}{r} \\[7ex] t = \dfrac{\ln \left(\dfrac{3P}{P}\right)}{0.09} \\[7ex] = \dfrac{\ln 3}{0.015} \\[5ex] = \dfrac{1.09861229}{0.015} \\[5ex] = 73.2408192 \\[3ex] \approx 73.24\:years \\[3ex]$ Ceteris paribus, it will take approximately $73.24\:years$ for the population of SamDom For Peace City to triple in size, if the population is growing at a rate of $1.5\%$ compounded continuously.
(3.) Nahum invested $10,000 in a bank that pays 13.7% compounded continuously. (a.) How much money will he have after 2 years? (b.) If another bank will pay Nahum 14% compounded quarterly, how much would he have after 2 years?$ (a.) \\[3ex] \underline{Continuous\:\:Compound\:\:Interest} \\[3ex] A = Pe^{rt} \\[3ex] P = \$10000 \\[3ex] t = 2 \: years \\[3ex] r = 13.7\% = \dfrac{13.7}{100} = 0.137 \\[5ex] A = 10000 * e^{0.137 * 2} \\[3ex] = 10000 * e^{0.274} \\[3ex] = 10000 * 1.315214802 \\[3ex] = 13152.14802 \\[3ex] A \approx \$13,152.15 \\[3ex] (b.) \\[3ex] \underline{Compound\:\:Interest} \\[3ex] A = P\left(1 + \dfrac{r}{m}\right)^{mt} \\[5ex] P = \$10000 \\[3ex] t = 2 \: years \\[3ex] m = 4 \\[3ex] r = 14\% = \dfrac{14}{100} = 0.14 \\[5ex] A = 10000 * \left(1 + \dfrac{0.14}{4}\right)^{4 * 2} \\[5ex] = 10000 * (1 + 0.035)^{8} \\[3ex] = 10000 * (1.035)^{8} \\[3ex] = 10000 * 1.316809037 \\[3ex] = 13168.09037 \\[3ex] A \approx \$13,168.09 $(4.) Matthew has$1000 to invest at 6% per annum compounded quarterly.
(a.) How long will it take before he has 1450$? (b.) If the compounding is continuous, how long will it be?$ (a.) \\[3ex] \underline{Compound\:\:Interest} \\[3ex] P = \$1000 \\[3ex] r = 6\% = \dfrac{6}{100} = 0.06 \\[5ex] A = \$1450 \\[3ex] Compounded\:\:quarterly \rightarrow m = 4 \\[3ex] t = \dfrac{\log \left(\dfrac{A}{P}\right)}{m\log \left(1 + \dfrac{r}{m}\right)} \\[7ex] t = \dfrac{\log \left(\dfrac{1450}{1000}\right)}{4 * \log \left(1 + \dfrac{0.06}{4}\right)} \\[7ex] t = \dfrac{\log (1.45)}{4 * \log (1 + 0.015)} \\[7ex] = \dfrac{\log (1.45)}{4 * \log (1.015)} \\[7ex] = \dfrac{0.161368002}{4 * 0.006466042} \\[7ex] = \dfrac{0.161368002}{0.025864169} \\[5ex] = 6.23905613 \\[3ex] t \approx 6.24\:years \\[3ex] (b.) \\[3ex] \underline{Continuous\:\:Compound\:\:Interest} \\[3ex] P = \$1000 \\[3ex] r = 6\% = \dfrac{6}{100} = 0.06 \\[5ex] A = \$1450 \\[3ex] t = \dfrac{\ln \left({\dfrac{A}{P}}\right)}{r} \\[7ex] t = \dfrac{\ln \left({\dfrac{1450}{1000}}\right)}{0.06} \\[7ex] = \dfrac{\ln {(1.45)}}{0.06} \\[5ex] = \dfrac{0.371563556}{0.06} \\[5ex] = 6.19272594 \\[3ex] t \approx 6.19\:years $(5.) Ezra invested$2000 at an interest rate, $k$ compounded continuously.
The fund amounted to $2504.65 in 5 years. (a.) Calculate the interest rate. (b.) Calculate the balance after 10 years. (c.) After how long will the fund be doubled? Round to the nearest hundredth as needed.$ \underline{Continuous\:\:Compound\:\:Interest} \\[3ex] (a.) \\[3ex] P = \$2000 \\[3ex] r = k \\[3ex] A = \$2504.65 \\[3ex] t = 5\:years \\[3ex] r = \dfrac{\ln \left({\dfrac{A}{P}}\right)}{t} \\[7ex] r = \dfrac{\ln \left({\dfrac{2504.65}{2000}}\right)}{5} \\[7ex] = \dfrac{\ln (1.252325)}{5} \\[5ex] = \dfrac{0.225001824}{5} \\[5ex] = 0.045000365 \\[3ex] to\:\:percent = 0.045000365(100) = 4.5000365 \\[3ex] r \approx 4.5\% \\[3ex] (b.) \\[3ex] After\: 10\: years; \\[3ex] t = 10\:years \\[3ex] A = ? \\[3ex] P = \$2000 \\[3ex] r = 0.045000365 \\[3ex] A = Pe^{rt} \\[5ex] A = 2000 * e^{0.045000365 * 10} \\[5ex] = 2000 * e^{0.45000365} \\[5ex] = 2000 * 1.568317907 \\[3ex] = 3136.635813 \\[3ex] A \approx \$3136.64 \\[3ex] (c.) \\[3ex] When\: the\: fund\: is\: doubled \\[3ex] P = \$2000 \: dollars \\[3ex] r = 0.045000365 \\[3ex] A = 2 * 2000 = \$4000 \\[3ex] t = ? \\[3ex] t = \dfrac{\ln \left({\dfrac{A}{P}}\right)}{r} \\[7ex] t = \dfrac{\ln \left({\dfrac{4000}{2000}}\right)}{0.045000365} \\[7ex] = \dfrac{\ln 2}{0.045000365} \\[5ex] = \dfrac{0.693147181}{0.045000365} \\[5ex] = 15.40314574 \\[3ex] t \approx 15.4\: years $(6.) Malachi invested a sum of$800 into an account that pays interest at the rate of 2.9% per year, compounded continuously.
How much money will be in the account after $8$ years?
Calculate the doubling time.
Round to the nearest hundredth as needed.

$\underline{Continuous\:\:Compound\:\:Interest} \\[3ex] P = \$800 \\[3ex] t = 8\: years \\[3ex] r = 2.9\% = \dfrac{2.9}{100} = 0.029 \\[5ex] A = Pe^{rt} \\[5ex] A = 800 * e^{0.029 * 8} \\[5ex] = 800 * e^{0.232} \\[5ex] = 800 * 1.261119729 \\[2ex] = 1008.895783 \\[3ex] A \approx \$1008.90 \\[3ex]$ Calculating the doubling time means "when" will the fund be doubled?
"How long" will it take for the money to be doubled?

$When\: the\: fund\: is\: doubled \\[3ex] P = \$800 \\[3ex] r = 0.029 \\[3ex] A = 2 * 800 = \$1600 \\[3ex] t = ? \\[2ex] t = \dfrac{\ln \left({\dfrac{A}{P}}\right)}{r} \\[7ex] t = \dfrac{\ln \left({\dfrac{1600}{800}}\right)}{0.029} \\[7ex] t = \dfrac{\ln 2}{0.029} \\[5ex] t = \dfrac{0.693147181}{0.0029} \\[5ex] t = 23.90162692 \\[3ex] t \approx 23.90\: years$
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