We want to simplify the process for calculating compounding because creating a table like the one on the previous page is time-consuming. Luckily, math is good at giving you ways to take shortcuts. To find an equation to represent this, if [latex]P_m[/latex] represents the amount of money after [latex]m[/latex] months, then we could write the recursive equation:
[latex]P_0 = $1000[/latex]
[latex]P_m= (1+0.0025)P_{m-1}[/latex]
You may recognize this as the recursive form of exponential growth.
recursive growth
Recall the underlying process of recursive growth. From a starting amount, [latex]P_0[/latex], each subsequent amount, [latex]P_m[/latex], grows in proportion to itself, [latex]P_{m-1}[/latex], at some rate [latex]r[/latex].
[latex]P_m=P_{m-1}+r\cdot P_{m-1}[/latex]
Factoring out the [latex]P_{m-1}[/latex] from each term on the right-hand side
[latex]P_m=(1+r)\cdot P_{m-1}[/latex].
In the example below, we’ll build an explicit equation for the growth.
Multiplying terms containing exponents
In the example below, you’ll need to use the rules for multiplying like bases containing exponents
[latex]a^{m}a^{n}=a^{m+n}[/latex]
That is, when multiplying like bases, we add the exponents.
Build an explicit equation for the growth of [latex]$1000[/latex] deposited in a bank account offering [latex]3\%[/latex] interest, compounded monthly.
Notice that the [latex]$1000[/latex] in the equation was [latex]P_0[/latex], the starting amount. We found [latex]1.0025[/latex] by adding one to the growth rate divided by [latex]12[/latex], since we were compounding [latex]12[/latex] times per year.
Note: This video uses [latex]k[/latex] as the variable instead of [latex]n[/latex].
While this formula works fine, it is more common to use a formula that involves the number of years, rather than the number of compounding periods. If [latex]t[/latex] is the number of years, then [latex]m = nt[/latex]. Making this change gives us the standard formula for compound interest.
How did we get [latex]m = nt[/latex]?
Recall that [latex]m[/latex] represents the number of compounding periods that an investment remains in the account, and [latex]n[/latex] represents the number of times per year that your interest is compounded. If your deposit earns interest compounded monthly, then [latex]n = 12[/latex]. If you leave the deposit in for [latex]1[/latex] year, then [latex]m = 12[/latex]. But if you leave the deposit in for [latex]2[/latex] years, then [latex]m = 2*12 = 24[/latex]. Looking at that another way, [latex]m = t(\text{years}) * n[/latex].
An investment of [latex]$1000[/latex] earning interest of [latex]4\%[/latex] compounded quarterly ([latex]4[/latex] times per year) is left in the account for [latex]3[/latex] years.
We have [latex]4[/latex] compounding periods per year, so [latex]n = 4[/latex].
If we leave our money in for [latex]1[/latex] year, the number of compounding periods is [latex]1*4: m=4[/latex].
If we leave our money in for [latex]3[/latex] years, [latex]m = 3*4[/latex], or [latex]12[/latex].
Knowing that investments are usually left to grow over years than over a number of compounding periods, we’ll adjust the formula slightly and just write [latex]nt[/latex]. This will make it easier to load the formula into a spreadsheet.
[latex]P_t[/latex] is the balance in the account after [latex]t[/latex] years.
[latex]P_0[/latex] is the starting balance of the account (also called initial deposit, or principal)
[latex]r[/latex] is the annual interest rate in decimal form
[latex]n[/latex] is the number of compounding periods in one year
If the compounding is done annually (once a year), [latex]n=1[/latex].
If the compounding is done quarterly, [latex]n=4[/latex].
If the compounding is done monthly, [latex]n=12[/latex].
If the compounding is done daily, [latex]n=365[/latex].
The most important thing to remember about using this formula is that it assumes that we put money in the account once and let it sit there earning interest.
In the next example, we show how to use the compound interest formula to find the balance on a certificate of deposit after [latex]20[/latex] years.
Don’t forget to convert percent to a decimal
Usually, in order to perform calculations on a number expressed in percent form, you’ll need to convert it to decimal form. The rate [latex]r[/latex] in interest formulas must be converted from percent to decimal form before you use the formula.
A certificate of deposit (CD) is a savings instrument that many banks offer. It usually gives a higher interest rate, but you cannot access your investment for a specified length of time. Suppose you deposit [latex]$3000[/latex] in a CD paying [latex]6\%[/latex] interest, compounded monthly. How much will you have in the account after [latex]20[/latex] years?
In this example,
[latex]P_0 = $3000[/latex]
the initial deposit
[latex]r = 0.06[/latex]
[latex]6\%[/latex] annual rate
[latex]n = 12[/latex]
[latex]12[/latex] months in [latex]1[/latex] year
[latex]t = 20[/latex]
since we’re looking for how much we’ll have after [latex]20[/latex] years
So [latex]{{P}_{20}}=3000{{\left(1+\frac{0.06}{12}\right)}^{20\times12}}=\$9930.61[/latex] (round your answer to the nearest penny)
A video walkthrough of this example problem is available below.
Note: This video uses [latex]k[/latex] for [latex]n[/latex], and [latex]N[/latex] for [latex]t[/latex].
Solving For Time
Note: This section assumes you’ve covered solving exponential equations using logarithms, either in prior classes or in the growth models module.
Often we are interested in how long it will take to accumulate money or how long we’d need to extend a loan to bring payments down to a reasonable level.
If you invest [latex]$2000[/latex] at [latex]6\%[/latex] compounded monthly, how long will it take the account to double in value?
This is a compound interest problem, since we are depositing money once and allowing it to grow. In this problem,
[latex]P_0 = $2000[/latex]
the initial deposit
[latex]r = 0.06[/latex]
[latex]6\%[/latex] annual rate
[latex]n = 12[/latex]
[latex]12[/latex] months in [latex]1[/latex] year
So our general equation is [latex]{{P}_{t}}=2000{{\left(1+\frac{0.06}{12}\right)}^{t\times12}}[/latex]. We also know that we want our ending amount to be double of [latex]$2000[/latex], which is [latex]$4000[/latex], so we’re looking for [latex]t[/latex] so that [latex]P_t = 4000[/latex]. To solve this, we set our equation for [latex]P_t[/latex] equal to [latex]4000[/latex].
[latex]\begin{array}{r@{\hfill}l} 4000=2000{{\left(1+\frac{0.06}{12}\right)}^{t\times12}} &&& \text{Divide both sides by } 2000 \\ 2={{\left(1.005\right)}^{12t}} &&& \text{To solve for the exponent, take the log of both sides} \\ \log\left(2\right)=\log\left({{\left(1.005\right)}^{12t}}\right) &&& \text{Use the exponent property of logs on the right side} \\ \log\left(2\right)=12t\log\left(1.005\right) &&& \text{Now we can divide both sides by } 12\text{log}(1.005) \\ \frac{\log\left(2\right)}{12\log\left(1.005\right)}=t &&& \text{Approximating this to a decimal} \\ t = 11.581 &&& \end{array}[/latex]
It will take about [latex]11.581[/latex] years for the account to double in value. Note that your answer may come out slightly differently if you had evaluated the logs to decimals and rounded during your calculations, but your answer should be close. For example if you rounded [latex]\text{log}(2)[/latex] to [latex]0.301[/latex] and [latex]\text{log}(1.005)[/latex] to [latex]0.00217[/latex], then your final answer would have been about [latex]11.577[/latex] years.
View this video for a walkthrough of this example.