The Central Idea
Compound interest applies a percentage growth factor repeatedly to the current balance.
If a principal \(P\) earns interest at rate \(r\) once per period, then after one period the balance is
\[ P(1+r). \]After two periods,
\[ P(1+r)^2. \]After \(t\) periods,
\[ \boxed{A=P(1+r)^t}. \]Compound interest is therefore a direct application of exponential growth.
Principal, Interest, and Balance
Financial models use several recurring terms.
- Principal is the initial amount invested or borrowed.
- Interest is the amount earned or charged for the use of money.
- Interest rate is the proportional rate used to calculate interest.
- Balance or accumulated amount is the total amount after interest has been applied.
In exponential notation, the principal plays the role of the initial value.
Simple Interest versus Compound Interest
Simple interest calculates interest from the original principal alone.
Compound interest calculates new interest from the current balance, which includes previously earned interest.
This distinction changes the mathematical structure:
\[ \text{simple interest}\longrightarrow\text{linear growth} \]while
\[ \boxed{\text{compound interest}\longrightarrow\text{exponential growth}}. \]Example: One Period of Interest
Suppose \(\$1000\) earns \(5\%\) interest during one year.
The interest earned is
\[ 1000(0.05)=50. \]The new balance is
\[ 1000+50=1050. \]Equivalently,
\[ 1000(1.05)=1050. \]Therefore,
\[ \boxed{A=1050}. \]Example: Two Periods of Compound Interest
Continue the previous example for another year.
The second year's interest is calculated from
\[ 1050, \]not from the original \(1000\).
Thus
\[ 1050(1.05)=1102.50. \]Equivalently,
\[ 1000(1.05)^2=1102.50. \]The interest itself has begun earning interest.
Compounding More Than Once per Year
Interest is often compounded several times during one year.
If the annual nominal rate is \(r\) and interest is compounded \(n\) times per year, then the rate applied during each compounding period is
\[ \frac{r}{n}. \]Over \(t\) years, the number of compounding periods is
\[ nt. \]The Periodic Compound Interest Formula
Combining the periodic rate and number of periods gives
\[ \boxed{ A=P\left(1+\frac{r}{n}\right)^{nt} }. \]In this formula:
- \(A\) is the accumulated amount;
- \(P\) is the principal;
- \(r\) is the nominal annual interest rate written as a decimal;
- \(n\) is the number of compounding periods per year;
- \(t\) is the time in years.
Why the Rate Is Divided by \(n\)
Suppose an account has a nominal annual rate of \(12\%\) compounded monthly.
Monthly compounding means
\[ n=12. \]The periodic rate is
\[ \frac{0.12}{12}=0.01. \]Therefore, each month the balance is multiplied by
\[ \boxed{1.01}. \]Why the Exponent Is \(nt\)
If interest is compounded \(n\) times per year for \(t\) years, then the total number of compounding events is
\[ n\times t. \]For monthly compounding over \(5\) years,
\[ n=12,\qquad t=5. \]Therefore, the number of compounding periods is
\[ 12(5)=60. \]The periodic growth factor is applied \(60\) times.
Common Compounding Frequencies
Some common values of \(n\) are:
\[ \begin{array}{c|c} \text{Compounding frequency} & n\\ \hline \text{annually} & 1\\ \text{semiannually} & 2\\ \text{quarterly} & 4\\ \text{monthly} & 12\\ \text{daily} & 365 \end{array} \]The exact convention used in a financial contract should always be checked, especially for daily compounding.
Example: Annual Compounding
Invest \(\$4000\) at \(6\%\) compounded annually for \(3\) years.
Here,
\[ P=4000,\qquad r=0.06,\qquad n=1,\qquad t=3. \]Therefore,
\[ A = 4000\left(1+\frac{0.06}{1}\right)^{1(3)}. \]Thus
\[ A=4000(1.06)^3. \]So
\[ \boxed{A\approx4764.06}. \]Example: Quarterly Compounding
Invest \(\$4000\) at \(6\%\) compounded quarterly for \(3\) years.
Here,
\[ n=4. \]The periodic rate is
\[ \frac{0.06}{4}=0.015. \]The number of periods is
\[ 4(3)=12. \]Therefore,
\[ A=4000(1.015)^{12}. \]Hence,
\[ \boxed{A\approx4782.47}. \]Example: Monthly Compounding
Invest \(\$4000\) at \(6\%\) compounded monthly for \(3\) years.
Here,
\[ n=12. \]Therefore,
\[ A = 4000\left(1+\frac{0.06}{12}\right)^{36}. \]So
\[ A=4000(1.005)^{36}. \]Therefore,
\[ \boxed{A\approx4786.72}. \]More Frequent Compounding
With the same principal, nominal annual rate, and elapsed time, more frequent compounding generally produces a slightly larger final amount.
This happens because interest is credited earlier and therefore begins earning additional interest sooner.
The increase from additional compounding frequency becomes progressively smaller as \(n\) becomes large.
Nominal Annual Rate
In
\[ A=P\left(1+\frac{r}{n}\right)^{nt}, \]the quantity \(r\) is typically a nominal annual rate.
The actual periodic rate applied during each compounding interval is
\[ \frac{r}{n}. \]Because interest compounds, the effective growth over a full year is generally not exactly \(r\).
Effective Annual Rate
The effective annual rate measures the actual percentage growth over one year after compounding is taken into account.
Starting with one unit of principal, the one-year accumulation factor is
\[ \left(1+\frac{r}{n}\right)^n. \]Therefore, the effective annual rate is
\[ \boxed{ r_{\text{eff}} = \left(1+\frac{r}{n}\right)^n-1 }. \]Example: Effective Annual Rate
Suppose a nominal rate of \(12\%\) is compounded monthly.
Then
\[ r=0.12,\qquad n=12. \]The effective annual rate is
\[ r_{\text{eff}} = \left(1+\frac{0.12}{12}\right)^{12}-1. \]Thus
\[ r_{\text{eff}} = (1.01)^{12}-1 \approx0.126825. \]Therefore, the effective annual rate is approximately
\[ \boxed{12.68\%}. \]Nominal Rate and Effective Rate Are Not the Same
A nominal annual rate describes the quoted annual rate before the effect of within-year compounding is incorporated.
The effective annual rate measures the actual one-year multiplicative growth.
When interest compounds more than once per year at a positive nominal rate, the effective annual rate is greater than the nominal annual rate.
Interest Earned
The accumulated amount \(A\) includes both the original principal and the interest earned.
Therefore,
\[ \boxed{\text{interest earned}=A-P}. \]If an investment grows from \(\$5000\) to \(\$5832\), then the interest earned is
\[ 5832-5000=832. \]Thus,
\[ \boxed{\$832} \]is interest.
Present Value
Sometimes the future amount is known and we want to determine the amount that must be invested now.
Starting with
\[ A=P\left(1+\frac{r}{n}\right)^{nt}, \]divide by the accumulation factor:
\[ \boxed{ P= \frac{A} {\left(1+\frac{r}{n}\right)^{nt}} }. \]This process is called discounting.
Example: Finding Present Value
How much must be invested now to have \(\$10{,}000\) in \(5\) years at \(4\%\) compounded annually?
Use
\[ P=\frac{10000}{(1.04)^5}. \]Therefore,
\[ \boxed{P\approx8219.27}. \]Approximately \(\$8{,}219.27\) must be invested initially.
Solving for Time Requires an Inverse Operation
Suppose
\[ A=P(1+r)^t. \]If \(A\), \(P\), and \(r\) are known but \(t\) is unknown, the variable is in the exponent.
Ordinary algebraic isolation eventually requires logarithms.
Later lessons will develop logarithms precisely as the inverse tools needed to solve equations of this type.
What Happens as Compounding Becomes More Frequent?
The periodic compound interest formula is
\[ A=P\left(1+\frac{r}{n}\right)^{nt}. \]As \(n\) increases, the periodic interest rate
\[ \frac{r}{n} \]becomes smaller, but it is applied more often.
These two effects compete in a structured way and approach a finite limit.
The Number \(e\)
The constant
\[ e\approx2.718281828\ldots \]arises naturally from repeated compounding.
One defining limit is
\[ \boxed{ e= \lim_{n\to\infty} \left(1+\frac1n\right)^n }. \]The number \(e\) is irrational and becomes the natural base for continuous exponential growth and decay.
From Periodic to Continuous Compounding
Begin with
\[ A=P\left(1+\frac{r}{n}\right)^{nt}. \]As the number of compounding periods per year increases without bound, the accumulated amount approaches
\[ \boxed{A=Pe^{rt}}. \]This formula models continuous compounding.
The Continuous Compounding Formula
For continuous compounding,
\[ \boxed{A=Pe^{rt}}. \]Here:
- \(A\) is the accumulated amount;
- \(P\) is the principal;
- \(r\) is the annual rate written as a decimal;
- \(t\) is time in years;
- \(e\) is the natural exponential base.
Example: Continuous Compounding
Invest \(\$5000\) at \(4\%\) continuously compounded for \(6\) years.
Here,
\[ P=5000,\qquad r=0.04,\qquad t=6. \]Therefore,
\[ A=5000e^{0.04(6)}. \]Thus
\[ A=5000e^{0.24}. \]Therefore,
\[ \boxed{A\approx6356.25}. \]Continuous Compounding Is Still Exponential Growth
The formula
\[ A=Pe^{rt} \]has the same basic exponential architecture as
\[ A=ab^t. \]The initial value is \(P\), and the exponential part
\[ e^{rt} \]controls the growth.
Continuous compounding does not create a new type of function. It gives a particularly natural exponential model.
Continuous Growth Factor over One Year
For
\[ A=Pe^{rt}, \]set \(t=1\):
\[ A=Pe^r. \]Therefore, the one-year multiplication factor is
\[ \boxed{e^r}. \]The corresponding effective annual rate is
\[ \boxed{e^r-1}. \]Example: Effective Rate under Continuous Compounding
Suppose the continuously compounded annual rate is \(5\%\).
Then
\[ r=0.05. \]The effective one-year rate is
\[ e^{0.05}-1. \]Numerically,
\[ e^{0.05}-1\approx0.051271. \]Therefore, the effective annual rate is approximately
\[ \boxed{5.13\%}. \]Comparing Annual and Continuous Compounding
Suppose \(\$1000\) earns \(10\%\) for one year.
Compounded annually:
\[ A=1000(1.10)=1100. \]Compounded continuously:
\[ A=1000e^{0.10}\approx1105.17. \]Continuous compounding produces the larger amount, but the difference is finite.
More Frequent Compounding Has Diminishing Gains
Moving from annual to monthly compounding can noticeably increase the final amount.
Moving from daily compounding to extremely frequent compounding produces a much smaller additional gain.
Continuous compounding represents the limiting accumulation associated with indefinitely increasing the compounding frequency under the standard model.
Continuous Compounding Is a Limit, Not Infinite Money
The phrase compounded continuously can sound as though the balance receives infinitely many full interest payments.
That is not what the formula means.
As the number of compounding periods increases, each periodic rate becomes correspondingly smaller.
The combined process approaches the finite value
\[ Pe^{rt}. \]Continuous Exponential Decay
The natural exponential form can also represent decay.
If \(k>0\), then
\[ \boxed{A=A_0e^{-kt}} \]decreases as \(t\) increases.
The negative exponent causes the exponential factor to shrink:
\[ e^{-kt}=\frac{1}{e^{kt}}. \]This form appears frequently in physical and scientific decay models.
Nominal Rate versus Continuous Rate
In
\[ A=Pe^{rt}, \]the parameter \(r\) is the continuously compounded annual rate.
Its effective annual rate is
\[ e^r-1. \]Therefore, a continuously compounded rate and an effective annual rate should not be treated as interchangeable numerical quantities.
Equivalent Annual Growth
Different compounding conventions can produce the same one-year growth factor.
If a periodically compounded investment has annual factor
\[ \left(1+\frac{r}{n}\right)^n, \]and a continuously compounded investment has factor
\[ e^k, \]then equivalent annual growth requires
\[ \boxed{ \left(1+\frac{r}{n}\right)^n=e^k }. \]Logarithms will later allow us to solve such relationships efficiently.
Present Value under Continuous Compounding
Starting with
\[ A=Pe^{rt}, \]divide by \(e^{rt}\):
\[ \boxed{P=Ae^{-rt}}. \]This gives the present value of a future amount under continuous compounding.
Example: Continuous Present Value
How much must be invested now to have \(\$20{,}000\) in \(8\) years at a continuously compounded rate of \(3\%\)?
Use
\[ P=20000e^{-0.03(8)}. \]Therefore,
\[ P=20000e^{-0.24}. \]Thus,
\[ \boxed{P\approx15732.56}. \]Financial Models Depend on Assumptions
Compound-interest formulas assume that the stated rate and compounding convention remain fixed throughout the modeled interval.
Actual financial products can include:
- changing interest rates;
- fees;
- taxes;
- minimum balances;
- additional deposits or withdrawals;
- different day-count conventions;
- contract-specific compounding rules.
The mathematical model must match the actual financial terms.
Compound Interest with No Additional Deposits
The formulas in this lesson model a single initial principal that grows without additional deposits or withdrawals.
Regular contributions create a different structure involving a sum of many exponential contributions.
Those models lead to annuities and other financial mathematics topics.
Common Error: Forgetting to Convert Percent to Decimal
A \(6\%\) rate must be entered as
\[ 0.06, \]not \(6\).
Thus quarterly compounding uses
\[ 1+\frac{0.06}{4}, \]not
\[ 1+\frac64. \]Common Error: Dividing the Principal by \(n\)
In
\[ A=P\left(1+\frac{r}{n}\right)^{nt}, \]the annual rate \(r\) is divided by \(n\).
The principal \(P\) is not divided by the compounding frequency.
The entire principal remains invested throughout the modeled period.
Common Error: Forgetting to Multiply the Exponent by \(n\)
If interest compounds monthly for \(4\) years, the number of periods is
\[ 12(4)=48. \]The exponent is therefore
\[ 48, \]not \(4\).
Common Error: Confusing Annual Rate with Periodic Rate
A nominal annual rate of \(8\%\) compounded quarterly does not mean that \(8\%\) is applied every quarter.
The quarterly rate is
\[ \frac{0.08}{4}=0.02. \]Thus each quarter uses growth factor
\[ \boxed{1.02}. \]Common Error: Confusing Amount with Interest Earned
If
\[ A=6500 \]and
\[ P=5000, \]then \(6500\) is the total accumulated balance.
The interest earned is
\[ 6500-5000=1500. \]Therefore,
\[ \boxed{\text{interest}=1500}. \]Common Error: Using the Periodic Formula for Continuous Compounding
Continuous compounding is modeled by
\[ \boxed{A=Pe^{rt}}, \]not by choosing an arbitrarily enormous value of \(n\) in a final exact formula.
Large values of \(n\) approximate the continuous limit, while \(Pe^{rt}\) represents that limit directly.
Common Error: Assuming More Frequent Compounding Changes the Nominal Rate
If the nominal annual rate remains \(6\%\), changing from quarterly to monthly compounding does not change \(r\) from \(0.06\).
What changes are:
\[ \frac{r}{n} \]and
\[ nt. \]The rate per period becomes smaller while the number of periods becomes larger.
A Reliable Procedure for Periodic Compounding
- Identify the principal \(P\).
- Convert the nominal annual rate to decimal form.
- Identify the number of compounding periods per year \(n\).
- Express time \(t\) in years.
- Compute the periodic rate \(r/n\).
- Compute the total number of periods \(nt\).
- Substitute into \(A=P(1+r/n)^{nt}\).
- Evaluate the exponential expression before rounding.
- Round money appropriately, usually to the nearest cent.
- If asked for interest earned, subtract \(P\) from \(A\).
A Reliable Procedure for Continuous Compounding
- Identify the principal \(P\).
- Convert the annual rate \(r\) to decimal form.
- Express time \(t\) in years.
- Substitute into \(A=Pe^{rt}\).
- Evaluate \(rt\).
- Evaluate the exponential factor \(e^{rt}\).
- Multiply by \(P\).
- Round only at the end unless the problem states otherwise.
Example: Full Periodic-Compounding Workflow
Invest \(\$7500\) at \(5.2\%\) compounded monthly for \(4\) years.
Identify the quantities:
\[ P=7500,\qquad r=0.052,\qquad n=12,\qquad t=4. \]Substitute:
\[ A= 7500\left(1+\frac{0.052}{12}\right)^{48}. \]Therefore,
\[ \boxed{A\approx9229.95}. \]The interest earned is approximately
\[ 9229.95-7500 = \boxed{1729.95}. \]Example: Full Continuous-Compounding Workflow
Invest \(\$7500\) at \(5.2\%\) compounded continuously for \(4\) years.
Use
\[ A=Pe^{rt}. \]Substitute:
\[ A=7500e^{0.052(4)}. \]Thus
\[ A=7500e^{0.208}. \]Therefore,
\[ \boxed{A\approx9234.10}. \]Practice 1: Annual Compounding
Find the balance of \(\$2000\) invested at \(5\%\) compounded annually for \(4\) years.
Practice 2: Quarterly Compounding
Find the balance of \(\$3000\) invested at \(8\%\) compounded quarterly for \(2\) years.
Practice 3: Monthly Compounding
Find the balance of \(\$5000\) invested at \(4.8\%\) compounded monthly for \(3\) years.
Practice 4: Interest Earned
An investment of \(\$6000\) grows to \(\$7240\).
How much interest was earned?
Practice 5: Effective Annual Rate
Find the effective annual rate for a nominal annual rate of \(6\%\) compounded monthly.
Practice 6: Continuous Compounding
Find the balance of \(\$2500\) invested at \(7\%\) compounded continuously for \(5\) years.
Practice 7: Continuous Effective Rate
Find the effective annual rate corresponding to a continuously compounded rate of \(4\%\).
Practice 8: Present Value
How much must be invested today to have \(\$15{,}000\) in \(6\) years at \(5\%\) compounded annually?
Practice 9: Continuous Present Value
How much must be invested today to have \(\$12{,}000\) in \(10\) years at \(3\%\) compounded continuously?
Practice 10: Choose the Formula
State which formula should be used in each case:
- \(6\%\) compounded monthly for \(8\) years.
- \(6\%\) compounded continuously for \(8\) years.
Practice Answers
Practice 1
\[ A=2000(1.05)^4. \]Therefore,
\[ \boxed{A\approx2431.01}. \]Practice 2
\[ A= 3000\left(1+\frac{0.08}{4}\right)^{4(2)} \] \[ =3000(1.02)^8. \]Therefore,
\[ \boxed{A\approx3514.98}. \]Practice 3
\[ A= 5000\left(1+\frac{0.048}{12}\right)^{36} \] \[ =5000(1.004)^{36}. \]Therefore,
\[ \boxed{A\approx5772.76}. \]Practice 4
\[ \text{interest}=7240-6000. \]Therefore,
\[ \boxed{\$1240}. \]Practice 5
\[ r_{\text{eff}} = \left(1+\frac{0.06}{12}\right)^{12}-1. \]Thus,
\[ r_{\text{eff}}\approx0.061678. \]Therefore, the effective annual rate is approximately
\[ \boxed{6.17\%}. \]Practice 6
\[ A=2500e^{0.07(5)} \] \[ =2500e^{0.35}. \]Therefore,
\[ \boxed{A\approx3547.67}. \]Practice 7
\[ r_{\text{eff}}=e^{0.04}-1. \]Therefore,
\[ r_{\text{eff}}\approx0.040811. \]The effective annual rate is approximately
\[ \boxed{4.08\%}. \]Practice 8
\[ P=\frac{15000}{(1.05)^6}. \]Therefore,
\[ \boxed{P\approx11193.23}. \]Practice 9
\[ P=12000e^{-0.03(10)} \] \[ =12000e^{-0.3}. \]Therefore,
\[ \boxed{P\approx8889.82}. \]Practice 10
For monthly compounding, use
\[ \boxed{ A=P\left(1+\frac{r}{n}\right)^{nt} }. \]For continuous compounding, use
\[ \boxed{A=Pe^{rt}}. \]What to Remember
Periodic compound interest is modeled by
\[ \boxed{ A=P\left(1+\frac{r}{n}\right)^{nt} }. \]Continuous compounding is modeled by
\[ \boxed{A=Pe^{rt}}. \]- \(P\) is the initial principal.
- \(r\) is an annual rate written as a decimal.
- \(n\) is the number of compounding periods per year.
- \(r/n\) is the periodic rate.
- \(nt\) is the total number of compounding periods.
- Compound interest is exponential because interest becomes part of the balance that earns later interest.
- More frequent compounding generally produces a larger accumulation at the same positive nominal rate.
- The effective annual rate accounts for within-year compounding.
- The constant \(e\) arises as the limiting base associated with increasingly frequent compounding.
- Interest earned is \(A-P\).
The central financial idea is: compound growth repeatedly applies interest to the current balance, not merely to the original principal.
Where We Go Next
Compound-interest formulas tell us how quickly a balance grows, but many exponential questions are more naturally phrased in terms of how long it takes a quantity to multiply or shrink by a particular factor.
The next lesson, Doubling Time and Half-Life, develops two of the most useful time scales in exponential mathematics and prepares the way for solving unknown exponents with logarithms.