Introduction
Interest is often described by asking how an account balance changes over time. In a continuous compounding model, the balance is represented by an exponential function. The formula is compact, but using it correctly depends on identifying what each quantity means, matching the units, and isolating the unknown carefully.
This tutorial focuses on using the model: calculating an amount from a principal, rate, and time, or working backward to find a rate or duration. The natural logarithm is needed for the latter calculations because it reverses the exponential function.
The Continuous-Compounding Model
The formula assumes that the principal and rate are constant during the time interval being modeled. For a positive initial principal, \(P>0\), the amount at time zero is \(A(0)=Pe^0=P\). The product \(rt\) has no units: if \(r\) is measured in inverse years and \(t\) in years, the units cancel.
A percentage rate must be converted to a decimal before it is substituted. For example, an annual rate of \(4\%\) is \(r=0.04\), not \(r=4\). The rate is annual, so \(t\) must be measured in years. A duration given in months must be converted by dividing by 12 before using the formula.
- \(P\): the balance at the start, sometimes called the principal.
- \(r\): the annual rate as a decimal. A positive value represents growth; a negative value represents decay.
- \(t\): the elapsed time in years, usually with \(t\geq 0\).
- \(A(t)\): the balance at time \(t\), in the same monetary units as \(P\).
For an account earning interest, the usual setting has \(P>0\) and \(r\geq 0\). The formula itself also represents exponential decay when \(r<0\). In either case, \(e^{rt}>0\), so a positive principal produces a positive amount.
Calculating an Amount
When \(P\), \(r\), and \(t\) are known, substitute them into \(A(t)=Pe^{rt}\). Evaluate the exponent first, then evaluate the exponential, and finally multiply by the principal. Keep the exact expression as long as practical and round the monetary result only at the end.
Worked Example: Find the balance after three years
An account begins with \(P=\$2{,}500\) and earns a continuous annual rate of \(4\%\). Find the balance after \(3\) years.
Convert the percentage to a decimal: \(r=4/100=0.04\). The given time is already in years, so \(t=3\). Substitute:
Rounded to the nearest cent, the balance is \(\$2{,}818.74\). Check by putting the unrounded amount and inputs back into the model: \(2500e^{(0.04)(3)}=2500e^{0.12}\approx \$2{,}818.74\), as calculated. The result exceeds the original principal, consistent with a positive rate over a positive time.
Solving for an Unknown Rate
Suppose the starting principal, final amount, and elapsed time are known, while the constant annual rate is unknown. Begin with \(A=Pe^{rt}\). Divide both sides by \(P\), then take the natural logarithm of both sides. This requires \(P>0\) and \(A>0\), which are natural conditions for positive account balances. If \(t>0\), divide by \(t\) to isolate \(r\).
The algebra also explains why the formula works. Dividing by \(P\) gives \(A/P=e^{rt}\). Since \(\ln\) reverses the exponential function on positive inputs, \(\ln(A/P)=rt\). Dividing by the nonzero time \(t\) yields the stated expression. Thus, for positive \(P\) and \(A\) and positive elapsed time, there is exactly one rate that fits the given data.
Worked Example: Find the annual rate
An account grows from \(\$1{,}000\) to \(\$1{,}210\) in \(2\) years. Find the continuous annual rate.
Here \(P=1000\), \(A=1210\), and \(t=2\). Apply the rate formula:
As a percentage, the rate is about \(9.531\%\) per year. Check by substituting the unrounded rate into the model: \(1000e^{(0.09531018)(2)}=1000e^{0.19062036}\approx1000(1.21)=\$1{,}210\). The computed rate reproduces the given final amount to the displayed precision.
Solving for an Unknown Duration
If the principal, rate, and target amount are known, take natural logarithms in the same way. The resulting equation is \(\ln(A/P)=rt\). When \(r\neq0\), divide by \(r\) to find the duration.
Worked Example: Find the time to reach a target
An account starts with \(\$800\) and earns a continuous annual rate of \(6\%\). How long will it take to reach \(\$1{,}000\)?
Use \(P=800\), \(A=1000\), and \(r=6/100=0.06\). Then:
The target is reached after about \(3.72\) years. Check by substituting the unrounded duration: \(800e^{(0.06)(3.7190592)}\approx800e^{0.22314355}=800(1.25)=\$1{,}000\). The positive time is consistent with growth from \(\$800\) to a larger target.
Finding the Principal and Checking Conditions
The model can also be rearranged to recover the initial principal when the amount, rate, and time are given. Since \(e^{rt}\) is positive, divide both sides by it: \(P=Ae^{-rt}=A/e^{rt}\). For instance, if an amount is \(\$1{,}500\) after \(2\) years at a rate of \(5\%\), then \(P=1500/e^{(0.05)(2)}=1500/e^{0.1}\approx \$1{,}357.26\). Substitution checks the result: \(1357.26e^{0.1}\approx \$1{,}500\), with the small difference caused by rounding.
Some inputs do not allow a unique solution for a requested rate or time. If \(t=0\), the equation \(A=Pe^{rt}\) reduces to \(A=P\), regardless of \(r\), so the rate cannot be identified from an observation made only at the initial time. If \(r=0\), then \(A=P\) at every time, so a duration cannot be determined from a target equal to \(P\); a different target is never reached under that model.
For \(t>0\), a positive \(P\), and a positive \(A\), the rate formula always gives a real rate. Its sign reflects the direction of change: if \(A>P\), then \(\ln(A/P)>0\) and \(r>0\); if \(A<P\), then \(r<0\); and if \(A=P\), then \(r=0\). For a duration with \(r>0\), a target \(A\geq P\) gives \(t\geq0\), while a target below \(P\) gives a negative value of \(t\), which describes a time before the starting point rather than a future duration. With \(r<0\), the direction is reversed.
Why the Logarithm Is Necessary
When the unknown is in the exponent, ordinary rearrangement cannot isolate it by dividing or subtracting. The logarithm provides the needed inverse operation. In this model, taking \(\ln\) changes the exponential equation \(A/P=e^{rt}\) into the linear equation \(\ln(A/P)=rt\). The formulas for rate and time follow by dividing this linear equation by the appropriate nonzero quantity.
This approach is useful whenever a constant exponential rate is used to describe a balance changing over time. It keeps the roles of the starting value, rate, and duration explicit, and it makes it possible to determine a missing parameter from observed values. The same algebraic discipline applies whether the model represents financial growth or exponential decrease.
Common Mistake: Entering a Percentage as a Whole Number
A frequent error is to substitute \(5\) for a rate of \(5\%\). For example, with \(P=\$1{,}000\), \(r=5\%\), and \(t=2\) years, the incorrect substitution \(1000e^{(5)(2)}=1000e^{10}\) gives an amount of about \(\$22{,}026{,}466\). This is not the model for a \(5\%\) annual rate: the percentage was not converted to a decimal.
The correction is \(5\%=5/100=0.05\). Therefore, \(A=1000e^{(0.05)(2)}=1000e^{0.1}\approx \$1{,}105.17\). Check the corrected answer by substitution: \(1000e^{0.1}\approx \$1{,}105.17\). Always convert a percentage to a decimal before using it as \(r\).
Summary
The continuous-compounding model is \(A(t)=Pe^{rt}\), with \(P\) as the initial principal, \(r\) as the annual decimal rate, \(t\) as time in years, and \(A(t)\) as the amount. Substitute known values to calculate an amount. To solve for an unknown rate or duration, take the natural logarithm after dividing by \(P\): \(\ln(A/P)=rt\). Check that the inputs satisfy the conditions for the calculation, keep units consistent, and verify the result in the original model.
Check Your Understanding
Answer these questions using the model and conditions discussed above.
- An account starts with \(\$600\) and earns \(3\%\) per year continuously for \(4\) years. What expression gives the final balance before rounding?
- If an amount grows from \(\$500\) to \(\$600\) in \(5\) years, what formula gives the annual rate as a decimal?
- Write an expression for the time needed for \(\$900\) to reach \(\$1{,}000\) at a continuous annual rate of \(4\%\).
- Why must a rate of \(7\%\) be entered as \(0.07\), rather than \(7\)?
- Can the rate be determined from \(P\), \(A\), and \(t\) if \(t=0\)? Explain using the model.
