The Central Idea
Interest depends on three basic quantities:
- the amount of money invested or borrowed;
- the interest rate;
- the amount of time.
The original amount is called the principal.
Simple interest is calculated only on the original principal.
Compound interest is calculated on the accumulated balance.
Principal, Interest, and Amount
Let
- \(P\) = principal;
- \(I\) = interest earned or charged;
- \(A\) = accumulated amount or future value.
These quantities satisfy
\[ \boxed{A=P+I}. \]Therefore,
\[ \boxed{I=A-P}. \]Interest Rates as Decimals
Interest formulas use rates in decimal form.
For example,
\[ 6\%=0.06, \] \[ 4.5\%=0.045, \]and
\[ 0.8\%=0.008. \]Divide a percent by \(100\) before substituting it into a formula.
Simple Interest
Simple interest is calculated on the original principal throughout the entire investment or loan.
If \(P\) is principal, \(r\) is the annual interest rate as a decimal, and \(t\) is time in years, then
\[ \boxed{I=Prt}. \]Simple-Interest Amount Formula
Since
\[ A=P+I \]and
\[ I=Prt, \]we have
\[ A=P+Prt. \]Factor \(P\):
\[ \boxed{A=P(1+rt)}. \]Example: Simple Interest Earned
Find the simple interest on \(\$2{,}500\) invested at \(6\%\) annually for \(4\) years.
\[ I=Prt. \] \[ I=2500(0.06)(4). \] \[ I=600. \]Therefore,
\[ \boxed{\$600\text{ interest}}. \]Example: Simple-Interest Amount
Continue the previous example.
\[ A=P+I. \] \[ A=2500+600. \] \[ A=3100. \]Therefore,
\[ \boxed{\$3{,}100}. \]Simple Interest Produces Linear Growth
For fixed \(P\) and \(r\),
\[ A=P(1+rt). \]Rewrite:
\[ A=P+(Pr)t. \]This is linear in \(t\).
The starting value is \(P\), and the amount added each year is
\[ \boxed{Pr}. \]Example: Annual Increase under Simple Interest
A principal of \(\$4{,}000\) earns \(5\%\) simple interest.
The annual interest is
\[ Pr=4000(0.05)=200. \]Therefore, the balance increases by
\[ \boxed{\$200\text{ each year}}. \]Solving Simple Interest for Principal
From
\[ I=Prt, \]solve for \(P\):
\[ \boxed{ P=\frac{I}{rt} }. \]Example: Find the Principal
An investment earns \(\$540\) in simple interest at \(4.5\%\) for \(3\) years.
\[ P= \frac{540}{0.045(3)}. \] \[ P= \frac{540}{0.135}. \] \[ P=4000. \]Therefore,
\[ \boxed{\$4{,}000}. \]Solving Simple Interest for Rate
From
\[ I=Prt, \]solve for \(r\):
\[ \boxed{ r=\frac{I}{Pt} }. \]Example: Find a Simple Interest Rate
A principal of \(\$5{,}000\) earns \(\$750\) in simple interest over \(5\) years.
\[ r= \frac{750}{5000(5)}. \] \[ r=0.03. \]Therefore,
\[ \boxed{3\%}. \]Solving Simple Interest for Time
From
\[ I=Prt, \]solve for \(t\):
\[ \boxed{ t=\frac{I}{Pr} }. \]Example: Find Simple-Interest Time
How long does \(\$3{,}000\) need to earn \(\$720\) at \(6\%\) simple interest?
\[ t= \frac{720}{3000(0.06)}. \] \[ t= \frac{720}{180}=4. \]Therefore,
\[ \boxed{4\text{ years}}. \]Time Units Must Match the Rate
If \(r\) is an annual interest rate, then \(t\) should be measured in years.
For example,
\[ 9\text{ months} = \frac9{12}\text{ year} = 0.75\text{ year}. \]Convert time units before using an annual-rate formula.
Example: Simple Interest for Months
Find the simple interest on \(\$8{,}000\) at \(7\%\) for \(9\) months.
Convert time:
\[ t=\frac9{12}=0.75. \]Then
\[ I=8000(0.07)(0.75). \] \[ I=420. \]Therefore,
\[ \boxed{\$420}. \]Compound Interest
Under compound interest, interest is periodically added to the balance.
Future interest is then calculated on that larger balance.
This produces repeated multiplicative growth rather than repeated additive growth.
One Compounding Period
Suppose the periodic interest rate is \(i\).
After one period,
\[ A=P+Pi. \]Factor:
\[ \boxed{A=P(1+i)}. \]Repeated Compounding
After two periods,
\[ A=P(1+i)^2. \]After three periods,
\[ A=P(1+i)^3. \]After \(N\) periods,
\[ \boxed{ A=P(1+i)^N }. \]Nominal Annual Rate with \(n\) Compounding Periods
Suppose the nominal annual interest rate is \(r\), compounded \(n\) times per year.
The periodic rate is
\[ \boxed{\frac{r}{n}}. \]Over \(t\) years, the number of compounding periods is
\[ \boxed{nt}. \]The Compound-Interest Formula
Combining the periodic rate and the number of periods gives
\[ \boxed{ A = P \left( 1+\frac{r}{n} \right)^{nt} }. \]Here:
- \(P\) = principal;
- \(r\) = nominal annual rate as a decimal;
- \(n\) = compounding periods per year;
- \(t\) = years;
- \(A\) = accumulated amount.
Common Compounding Frequencies
Typical values of \(n\) include:
\[ \begin{array}{c|c} \text{compounding frequency} & n\\ \hline \text{annual} & 1\\ \text{semiannual} & 2\\ \text{quarterly} & 4\\ \text{monthly} & 12\\ \text{daily} & 365 \end{array} \]Use the frequency stated in the problem.
Example: Annual Compounding
Invest \(\$5{,}000\) at \(6\%\) compounded annually for \(4\) years.
\[ A= 5000(1.06)^4. \] \[ A\approx6312.38. \]Therefore,
\[ \boxed{\$6{,}312.38\text{ approximately}}. \]Example: Quarterly Compounding
Invest \(\$5{,}000\) at a nominal annual rate of \(6\%\), compounded quarterly for \(4\) years.
Here,
\[ n=4. \]Periodic rate:
\[ \frac{0.06}{4}=0.015. \]Number of periods:
\[ 4(4)=16. \]Therefore,
\[ A= 5000(1.015)^{16}. \] \[ A\approx6344.93. \]Thus,
\[ \boxed{\$6{,}344.93\text{ approximately}}. \]Interest Earned under Compounding
Once the final amount is known, compound interest earned is
\[ \boxed{I=A-P}. \]In the quarterly example,
\[ I\approx6344.93-5000. \] \[ I\approx1344.93. \]Therefore,
\[ \boxed{\$1{,}344.93\text{ interest approximately}}. \]Compound Growth Is Exponential
In
\[ A=P(1+i)^N, \]the number of periods \(N\) appears in the exponent.
Each period multiplies the current balance by the same growth factor
\[ \boxed{1+i}. \]This is the defining structure of exponential growth.
Simple versus Compound Interest
Simple interest:
\[ A=P(1+rt). \]Compound interest:
\[ A= P \left( 1+\frac{r}{n} \right)^{nt}. \]Simple interest grows by equal additions.
Compound interest grows by equal multiplicative factors.
Example: Compare Simple and Compound Interest
Compare \(\$10{,}000\) at \(5\%\) for \(10\) years.
Simple interest:
\[ A_s=10000(1+0.05(10)). \] \[ A_s=15000. \]Annual compounding:
\[ A_c=10000(1.05)^{10}. \] \[ A_c\approx16288.95. \]Therefore, annual compounding produces approximately
\[ 16288.95-15000=1288.95 \]more than simple interest over the same period.
Present Value
Sometimes the future amount \(A\) is known and the required initial principal \(P\) is unknown.
Start with
\[ A= P \left( 1+\frac{r}{n} \right)^{nt}. \]Divide:
\[ \boxed{ P = \frac{A}{ \left( 1+\frac{r}{n} \right)^{nt} } }. \]Example: Find Present Value
How much must be invested now to have \(\$10{,}000\) in \(5\) years at \(4\%\) compounded annually?
\[ P= \frac{10000}{(1.04)^5}. \] \[ P\approx8219.27. \]Therefore,
\[ \boxed{\$8{,}219.27\text{ approximately}}. \]Solving Compound Interest for Rate
With annual compounding,
\[ A=P(1+r)^t. \]Divide by \(P\):
\[ \frac{A}{P}=(1+r)^t. \]Take the \(t\)-th root:
\[ 1+r= \left( \frac{A}{P} \right)^{1/t}. \]Therefore,
\[ \boxed{ r= \left( \frac{A}{P} \right)^{1/t}-1 }. \]Example: Find an Annual Compound Rate
An investment grows from \(\$4{,}000\) to \(\$5{,}324\) in \(3\) years with annual compounding.
\[ r= \left( \frac{5324}{4000} \right)^{1/3}-1. \]Since
\[ \frac{5324}{4000}=1.331, \]and
\[ 1.331=1.1^3, \]we obtain
\[ r=0.10. \]Therefore,
\[ \boxed{10\%}. \]Solving Compound Interest for Time
With annual compounding,
\[ A=P(1+r)^t. \]Divide:
\[ \frac{A}{P}=(1+r)^t. \]Take logarithms:
\[ \ln\left(\frac{A}{P}\right) = t\ln(1+r). \]Therefore,
\[ \boxed{ t= \frac{ \ln(A/P) }{ \ln(1+r) } }. \]Example: Find Compound-Growth Time
How long does \(\$5{,}000\) take to grow to \(\$8{,}000\) at \(6\%\) compounded annually?
\[ t= \frac{\ln(8000/5000)}{\ln(1.06)}. \] \[ t\approx8.06. \]Therefore,
\[ \boxed{8.06\text{ years approximately}}. \]Compounding More Frequently
For the same nominal annual rate \(r\), increasing the compounding frequency generally increases the accumulated amount.
Annual compounding uses
\[ n=1, \]while monthly compounding uses
\[ n=12. \]The increase becomes smaller as compounding becomes more frequent.
Effective Annual Rate
A nominal annual rate does not by itself describe the exact one-year growth when interest is compounded more than once per year.
The effective annual rate is
\[ \boxed{ r_{\text{eff}} = \left( 1+\frac{r}{n} \right)^n-1 }. \]Example: Effective Annual Rate
Find the effective annual rate for a nominal \(6\%\) rate compounded monthly.
\[ r_{\text{eff}} = \left( 1+\frac{0.06}{12} \right)^{12}-1. \] \[ r_{\text{eff}} = (1.005)^{12}-1. \] \[ r_{\text{eff}}\approx0.061678. \]Therefore,
\[ \boxed{6.17\%\text{ approximately}}. \]Nominal versus Effective Rate
A nominal rate is the stated annual rate used to determine the periodic rate.
An effective annual rate is the actual one-year percentage growth after compounding.
When compounding occurs more than once per year at a positive rate,
\[ r_{\text{eff}}>r. \]Continuous Compounding
As compounding becomes increasingly frequent, the compound-interest model approaches continuous compounding.
The continuously compounded amount is
\[ \boxed{ A=Pe^{rt} }. \]Here \(e\) is the natural exponential base.
Example: Continuous Compounding
Invest \(\$7{,}500\) at \(5\%\) continuously compounded for \(6\) years.
\[ A=7500e^{0.05(6)}. \] \[ A=7500e^{0.30}. \] \[ A\approx10123.94. \]Therefore,
\[ \boxed{\$10{,}123.94\text{ approximately}}. \]Present Value under Continuous Compounding
From
\[ A=Pe^{rt}, \]solve for \(P\):
\[ \boxed{ P=Ae^{-rt} }. \]Example: Continuous Present Value
How much must be invested now to reach \(\$20{,}000\) in \(8\) years at \(4\%\) continuously compounded?
\[ P=20000e^{-0.04(8)}. \] \[ P=20000e^{-0.32}. \] \[ P\approx14522.98. \]Therefore,
\[ \boxed{\$14{,}522.98\text{ approximately}}. \]Solving Continuous Compounding for Time
Start with
\[ A=Pe^{rt}. \]Divide:
\[ \frac{A}{P}=e^{rt}. \]Take natural logarithms:
\[ \ln\left(\frac{A}{P}\right)=rt. \]Therefore,
\[ \boxed{ t= \frac{ \ln(A/P) }{r} }. \]Solving Continuous Compounding for Rate
From
\[ \ln\left(\frac{A}{P}\right)=rt, \]solve for \(r\):
\[ \boxed{ r= \frac{ \ln(A/P) }{t} }. \]Doubling Time under Continuous Compounding
If an amount doubles,
\[ A=2P. \]Using
\[ 2P=Pe^{rt}, \]cancel \(P\):
\[ 2=e^{rt}. \]Therefore,
\[ \boxed{ t=\frac{\ln2}{r} }. \]Example: Continuous Doubling Time
Find the doubling time at \(7\%\) continuous compounding.
\[ t= \frac{\ln2}{0.07}. \] \[ t\approx9.90. \]Therefore,
\[ \boxed{9.90\text{ years approximately}}. \]Growth Factor View
Compound-interest models are easiest to understand through growth factors.
For periodic compounding, the growth factor per period is
\[ \boxed{ 1+\frac{r}{n} }. \]After \(nt\) periods, that factor has been applied \(nt\) times.
Interest Rate versus Growth Factor
If the periodic rate is \(2\%\), the growth factor is
\[ 1+0.02=1.02. \]A growth factor of \(1.02\) means the new balance is \(102\%\) of the previous balance after each period.
Compounding Frequency and Periodic Rate
A nominal annual rate of \(12\%\) compounded monthly does not mean \(12\%\) is applied every month.
The monthly periodic rate is
\[ \frac{0.12}{12}=0.01. \]Therefore, the monthly growth factor is
\[ \boxed{1.01}. \]Number of Periods
A nominal annual rate compounded monthly for \(3\) years uses
\[ 12(3)=36 \]compounding periods.
The exponent must count periods, not years alone.
Example: Monthly Compounding
Find the value of \(\$6{,}000\) at \(4.8\%\) compounded monthly for \(5\) years.
Here,
\[ \frac{r}{n} = \frac{0.048}{12} = 0.004, \]and
\[ nt=12(5)=60. \]Therefore,
\[ A=6000(1.004)^{60}. \] \[ A\approx7623.84. \]Thus,
\[ \boxed{\$7{,}623.84\text{ approximately}}. \]Comparing Compounding Frequencies
Suppose \(\$1{,}000\) earns a nominal \(8\%\) for one year.
Annual compounding:
\[ 1000(1.08)=1080. \]Quarterly compounding:
\[ 1000\left(1+\frac{0.08}{4}\right)^4. \]Monthly compounding:
\[ 1000\left(1+\frac{0.08}{12}\right)^{12}. \]Continuous compounding:
\[ 1000e^{0.08}. \]More frequent compounding increases the amount, but the increments become progressively smaller.
Simple Interest as a Rate Problem
Simple interest can be viewed as a constant-dollar rate problem.
The annual dollar interest rate is
\[ Pr. \]Therefore,
\[ I=(Pr)t. \]This is structurally similar to other rate-times-time models studied earlier.
Compound Interest as a Recursive Process
If \(A_k\) is the balance after period \(k\), then
\[ A_{k+1}=A_k(1+i). \]Starting with
\[ A_0=P, \]repeated substitution gives
\[ A_k=P(1+i)^k. \]Compound interest is therefore both an exponential model and a recurrence.
Checking Simple Interest
Under simple interest, equal time intervals should produce equal dollar interest amounts.
If a \(\$2{,}000\) principal earns \(5\%\) simple interest, each year adds
\[ 2000(0.05)=100. \]After \(3\) years, interest should be
\[ 3(100)=300. \]Checking Compound Interest
Under compound interest at a positive rate, the dollar amount of interest added each period generally increases because the balance grows.
A result showing the same dollar interest every period would describe simple interest rather than compound interest.
Checking with Magnitude
At a positive interest rate and positive time,
\[ A>P. \]If a compound-interest calculation produces an accumulated value smaller than the principal without withdrawals or negative rates, recheck the setup.
Common Error: Using Percent Instead of Decimal
A rate of \(6\%\) should be entered as
\[ 0.06, \]not
\[ 6. \]Common Error: Confusing Interest with Final Amount
In simple interest,
\[ I=Prt \]gives only the interest.
The final amount is
\[ \boxed{A=P+I}. \]Common Error: Forgetting to Divide the Rate by \(n\)
A nominal annual rate compounded monthly must be divided by \(12\) before being applied each month.
The periodic rate is
\[ \boxed{\frac{r}{12}}. \]Common Error: Forgetting to Multiply Time by \(n\)
Monthly compounding for \(5\) years means
\[ 12(5)=60 \]periods.
The exponent is \(60\), not \(5\).
Common Error: Applying Simple Interest to a Compound Problem
The formulas
\[ A=P(1+rt) \]and
\[ A= P \left( 1+\frac{r}{n} \right)^{nt} \]describe fundamentally different growth rules.
Read the problem carefully for words such as “simple,” “compounded monthly,” or “compounded continuously.”
Common Error: Treating Nominal Rate as Effective Rate
A stated nominal rate of \(6\%\) compounded monthly produces an effective annual rate slightly greater than \(6\%\).
Do not treat those two percentages as interchangeable.
Common Error: Rounding Too Early
Compound-interest calculations may involve many repeated multiplications.
Rounding intermediate values too aggressively can noticeably change the final result.
Keep extra calculator digits until the final monetary answer.
Common Error: Ignoring Time Units
An annual rate paired with \(18\) months should use
\[ t=1.5\text{ years} \]unless the formula has been rewritten in monthly terms.
A Reliable Procedure for Simple Interest
- Identify principal \(P\).
- Convert the annual rate to decimal form.
- Convert time to years if needed.
- Use \(I=Prt\).
- If final amount is requested, compute \(A=P+I\).
- Check units and reasonableness.
A Reliable Procedure for Periodic Compound Interest
- Identify \(P\), \(r\), \(n\), and \(t\).
- Convert \(r\) to decimal form.
- Compute the periodic rate \(r/n\).
- Compute the number of periods \(nt\).
- Use \(A=P(1+r/n)^{nt}\).
- Subtract principal if interest earned is requested.
- Round only at the end.
A Reliable Procedure for Unknowns in Compound Interest
- Write the correct compound-growth equation.
- Isolate the unknown algebraically when possible.
- Use roots when solving for a growth rate.
- Use logarithms when the unknown appears in an exponent.
- Check that the resulting rate or time is positive and contextually reasonable.
- Substitute back into the original model.
A Reliable Procedure for Continuous Compounding
- Use \(A=Pe^{rt}\).
- Convert the annual rate to decimal form.
- Use time in years.
- Apply logarithms when solving for \(r\) or \(t\).
- Check the result against the starting principal and expected growth direction.
Practice 1: Simple Interest
Find the simple interest on \(\$3{,}500\) at \(5\%\) for \(4\) years.
Practice 2: Simple-Interest Amount
Find the final amount for Practice 1.
Practice 3: Simple Interest with Months
Find the simple interest on \(\$6{,}000\) at \(8\%\) for \(9\) months.
Practice 4: Find Simple-Interest Principal
An investment earns \(\$900\) simple interest at \(6\%\) for \(5\) years. Find the principal.
Practice 5: Find Simple-Interest Rate
A \(\$4{,}000\) principal earns \(\$640\) simple interest in \(4\) years. Find the annual rate.
Practice 6: Annual Compound Interest
Find the value of \(\$5{,}000\) invested at \(7\%\) compounded annually for \(3\) years.
Practice 7: Monthly Compound Interest
Find the value of \(\$8{,}000\) at \(6\%\) compounded monthly for \(2\) years.
Practice 8: Compound Interest Earned
Using Practice 7, find the interest earned.
Practice 9: Present Value
How much must be invested now to have \(\$15{,}000\) in \(6\) years at \(5\%\) compounded annually?
Practice 10: Effective Annual Rate
Find the effective annual rate for a nominal \(8\%\) rate compounded quarterly.
Practice 11: Continuous Compounding
Find the value of \(\$10{,}000\) at \(4\%\) continuously compounded for \(5\) years.
Practice 12: Continuous Present Value
How much must be invested now to have \(\$25{,}000\) in \(10\) years at \(3\%\) continuously compounded?
Practice 13: Find Annual Compound Rate
An investment grows from \(\$5{,}000\) to \(\$6{,}655\) in \(3\) years with annual compounding. Find the rate.
Practice 14: Find Compound-Growth Time
How long does \(\$4{,}000\) take to grow to \(\$6{,}000\) at \(5\%\) compounded annually?
Practice 15: Identify the Error
A student evaluates a \(12\%\) nominal rate compounded monthly by using
\[ A=P(1.12)^{12t}. \]Explain the error and give the correct monthly-compounding model.
Practice Answers
Practice 1
\[ I=3500(0.05)(4). \] \[ I=700. \]Therefore,
\[ \boxed{\$700}. \]Practice 2
\[ A=3500+700. \] \[ \boxed{\$4{,}200}. \]Practice 3
Convert time:
\[ t=\frac9{12}=0.75. \] \[ I=6000(0.08)(0.75). \] \[ I=360. \]Therefore,
\[ \boxed{\$360}. \]Practice 4
\[ P= \frac{900}{0.06(5)}. \] \[ P= \frac{900}{0.30}=3000. \] \[ \boxed{\$3{,}000}. \]Practice 5
\[ r= \frac{640}{4000(4)}. \] \[ r=0.04. \] \[ \boxed{4\%}. \]Practice 6
\[ A=5000(1.07)^3. \] \[ A=5000(1.225043). \] \[ A=6125.215. \]Therefore,
\[ \boxed{\$6{,}125.22\text{ approximately}}. \]Practice 7
\[ A= 8000 \left( 1+\frac{0.06}{12} \right)^{24}. \] \[ A=8000(1.005)^{24}. \] \[ A\approx9017.28. \]Therefore,
\[ \boxed{\$9{,}017.28\text{ approximately}}. \]Practice 8
\[ I=A-P. \] \[ I\approx9017.28-8000. \] \[ \boxed{\$1{,}017.28\text{ approximately}}. \]Practice 9
\[ P= \frac{15000}{(1.05)^6}. \] \[ P\approx11193.23. \]Therefore,
\[ \boxed{\$11{,}193.23\text{ approximately}}. \]Practice 10
\[ r_{\text{eff}} = \left( 1+\frac{0.08}{4} \right)^4-1. \] \[ r_{\text{eff}} = (1.02)^4-1. \] \[ r_{\text{eff}}\approx0.082432. \]Therefore,
\[ \boxed{8.24\%\text{ approximately}}. \]Practice 11
\[ A=10000e^{0.04(5)}. \] \[ A=10000e^{0.20}. \] \[ A\approx12214.03. \]Therefore,
\[ \boxed{\$12{,}214.03\text{ approximately}}. \]Practice 12
\[ P=25000e^{-0.03(10)}. \] \[ P=25000e^{-0.30}. \] \[ P\approx18520.46. \]Therefore,
\[ \boxed{\$18{,}520.46\text{ approximately}}. \]Practice 13
\[ r= \left( \frac{6655}{5000} \right)^{1/3}-1. \] \[ \frac{6655}{5000}=1.331. \] \[ 1.331=1.1^3. \] \[ r=0.10. \]Therefore,
\[ \boxed{10\%}. \]Practice 14
\[ t= \frac{\ln(6000/4000)}{\ln(1.05)}. \] \[ t= \frac{\ln(1.5)}{\ln(1.05)}. \] \[ t\approx8.31. \]Therefore,
\[ \boxed{8.31\text{ years approximately}}. \]Practice 15
The student applies the full annual rate every month. The monthly periodic rate is
\[ \frac{0.12}{12}=0.01. \]Therefore, the correct model is
\[ \boxed{ A=P(1.01)^{12t} }. \]What to Remember
Simple interest is
\[ \boxed{I=Prt}. \]Its accumulated amount is
\[ \boxed{A=P(1+rt)}. \]Periodic compound interest is
\[ \boxed{ A= P \left( 1+\frac{r}{n} \right)^{nt} }. \]Continuous compounding is
\[ \boxed{A=Pe^{rt}}. \]Simple interest is linear in time; compound interest is exponential.
The most important habit is to identify the interest model and compounding frequency before choosing the formula.
Where We Go Next
Interest models are examples of broader growth processes. The same exponential structure appears in population growth, depreciation, decay, and many applied models in science and economics.
The next lesson, Growth, Depreciation, and Population Models, extends these ideas beyond financial interest.