2.2d Which Formula Do I Use?
When presented with a finance problem (on an exam or in real life), you're usually not told what type of problem it is or which equation to use. Here are some hints on deciding which equation to use based on the wording of the problem.
The easiest types of problem to identify are loans. Loan problems almost always include words like: "loan", "amortize" (the fancy word for loans), "finance (a car)", or "mortgage" (a home loan). Look for these words. If they're there, you're probably looking at a loan problem. To make sure, see if you're given what your monthly (or annual) payment is, or if you're trying to find a monthly payment.
If the problem is not a loan, the next question you want to ask is: "Am I putting money in an account and letting it sit, or am I making regular (monthly/annually/quarterly) payments or withdrawals?" If you're letting the money sit in the account with nothing but interest changing the balance, then you're looking at a compound interest problem. The exception would be bonds and other investments where the interest is not reinvested; in those cases you're looking at simple interest.
If you're making regular payments or withdrawals, the next questions is: "Am I putting money into the account, or am I pulling money out?" If you're putting money into the account on a regular basis (monthly/annually/quarterly) then you're looking at a basic Annuity problem. Basic annuities are when you are saving money. Usually in an annuity problem, your account starts empty, and has money in the future.
If you're pulling money out of the account on a regular basis, then you're looking at a Payout Annuity problem. Payout annuities are used for things like retirement income, where you start with money in your account, pull money out on a regular basis, and your account ends up empty in the future.
Remember, the most important part of answering any kind of question, money or otherwise, is first to correctly identify what the question is really asking, and to determine what approach will best allow you to solve the problem.
Note: This section assumes you've covered solving exponential equations using logarithms, either in prior classes or in the growth models chapter.
This is a compound interest problem, since we are depositing money once and allowing it to grow. In this problem,
So our general equation is
It will take about 11.581 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 log(2) to 0.301 and log(1.005) to 0.00217, then your final answer would have been about 11.577 years.
This is a savings annuity problem since we are making regular deposits into the account.
We don't know
N, but we want P_{N} to be $10,000.
Putting this into the equation:
Simplifying the fractions a bit
We want to isolate the exponential term, 1.0025
^{12N}, so multiply both sides by 0.0025
Divide both sides by 100
Add 1 to both sides
Now take the log of both sides
Use the exponent property of logs
Divide by 12log(1.0025)
Approximating to a decimal,
It will take about 7.447 years to grow the account to $10,000.
David Lippman, Math in Society, "Finance," licensed under a CC BYSA 3.0 license.
The easiest types of problem to identify are loans. Loan problems almost always include words like: "loan", "amortize" (the fancy word for loans), "finance (a car)", or "mortgage" (a home loan). Look for these words. If they're there, you're probably looking at a loan problem. To make sure, see if you're given what your monthly (or annual) payment is, or if you're trying to find a monthly payment.
If the problem is not a loan, the next question you want to ask is: "Am I putting money in an account and letting it sit, or am I making regular (monthly/annually/quarterly) payments or withdrawals?" If you're letting the money sit in the account with nothing but interest changing the balance, then you're looking at a compound interest problem. The exception would be bonds and other investments where the interest is not reinvested; in those cases you're looking at simple interest.
If you're making regular payments or withdrawals, the next questions is: "Am I putting money into the account, or am I pulling money out?" If you're putting money into the account on a regular basis (monthly/annually/quarterly) then you're looking at a basic Annuity problem. Basic annuities are when you are saving money. Usually in an annuity problem, your account starts empty, and has money in the future.
If you're pulling money out of the account on a regular basis, then you're looking at a Payout Annuity problem. Payout annuities are used for things like retirement income, where you start with money in your account, pull money out on a regular basis, and your account ends up empty in the future.
Remember, the most important part of answering any kind of question, money or otherwise, is first to correctly identify what the question is really asking, and to determine what approach will best allow you to solve the problem.
Try it Now
For each of the following scenarios, determine if it is a compound interest problem, a savings annuity problem, a payout annuity problem, or a loans problem. Then solve each problem. Marcy received an inheritance of $20,000, and invested it at 6% interest. She is going to use it for college, withdrawing money for tuition and expenses each quarter. How much can she take out each quarter if she has 3 years of school left?
 Paul wants to buy a new car. Rather than take out a loan, he decides to save $200 a month in an account earning 3% interest compounded monthly. How much will he have saved up after 3 years?
 Keisha is managing investments for a nonprofit company. They want to invest some money in an account earning 5% interest compounded annually with the goal to have $30,000 in the account in 6 years. How much should Keisha deposit into the account?
 Miao is going to finance new office equipment at a 2% rate over a 4 year term. If she can afford monthly payments of $100, how much new equipment can she buy?
 How much would you need to save every month in an account earning 4% interest to have $5,000 saved up in two years?
Solving for Time
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.Note: This section assumes you've covered solving exponential equations using logarithms, either in prior classes or in the growth models chapter.
Example 1
If you invest $2000 at 6% 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,
Initial deposit  $\displaystyle{P}_{{0}}=\${2000}$ 
6% annual rate  $\displaystyle{r}={0.06}$ 
12 months in 1 year  $\displaystyle{k}={12}$ 
$\displaystyle{P}_{{N}}={2000}{\left({1}+\frac{{0.06}}{{12}}\right)}^{{{N}\times{12}}}$
. We also know that we want our ending amount to be double of $2000, which is $4000, so we're looking for N so that P_{N} = 4000. To solve this, we set our equation for P_{N} equal to 4000.$\displaystyle{4000}={2000}{\left({1}+\frac{{0.06}}{{12}}\right)}^{{{N}\times{12}}}$ 

Divide both sides by 2000  $\displaystyle{2}={\left({1.005}\right)}^{{{12}{N}}}$ 
To solve for the exponent, take the log of both sides  $\displaystyle{\log{{\left({2}\right)}}}={\log{{\left({\left({1.005}\right)}^{{{12}{N}}}\right)}}}$ 
Use the exponent property of logs on the right side  $\displaystyle{\log{{\left({2}\right)}}}={12}{N}{\log{{\left({1.005}\right)}}}$ 
Now we can divide both sides by 12log(1.005)  $\displaystyle\frac{{{\log{{\left({2}\right)}}}}}{{{12}{\log{{\left({1.005}\right)}}}}}={N}$ 
Approximating this to a decimal  $\displaystyle{N}={11.581}$ 
Example 2
If you invest $100 each month into an account earning 3% compounded monthly, how long will it take the account to grow to $10,000?This is a savings annuity problem since we are making regular deposits into the account.
Monthly deposit  $\displaystyle{d}=\${100}$ 
3% annual rate  $\displaystyle{r}={0.03}$ 
12 months in 1 year  $\displaystyle{k}={12}$ 
N, but we want P_{N} to be $10,000.
Putting this into the equation:
$\displaystyle{10},{000}=\frac{{{100}{\left({\left({1}+\frac{{0.03}}{{12}}\right)}^{{{N}{\left({12}\right)}}}{1}\right)}}}{{{\left(\frac{{0.03}}{{12}}\right)}}}$
Simplifying the fractions a bit
$\displaystyle{10},{000}=\frac{{{100}{\left({\left({1.0025}\right)}^{{{12}{N}}}{1}\right)}}}{{0.0025}}$
We want to isolate the exponential term, 1.0025
^{12N}, so multiply both sides by 0.0025
$\displaystyle{25}={100}{\left({\left({1.0025}\right)}^{{{12}{N}}}{1}\right)}$
Divide both sides by 100
$\displaystyle{0.25}={\left({1.0025}\right)}^{{{12}{N}}}{1}$
Add 1 to both sides
$\displaystyle{1.25}={\left({1.0025}\right)}^{{{12}{N}}}$
Now take the log of both sides
$\displaystyle{\log{{\left({1.25}\right)}}}={\log{{\left({\left({1.0025}\right)}^{{{12}{N}}}\right)}}}$
Use the exponent property of logs
$\displaystyle{\log{{\left({1.25}\right)}}}={12}{N}{\log{{\left({1.0025}\right)}}}$
Divide by 12log(1.0025)
$\displaystyle\frac{{{\log{{\left({1.25}\right)}}}}}{{{12}{\log{{\left({1.0025}\right)}}}}}={N}$
Approximating to a decimal,
$\displaystyle{N}={7.447}$
yearsIt will take about 7.447 years to grow the account to $10,000.
Try it Now 6
Joel is considering putting a $1,000 laptop purchase on his credit card, which has an interest rate of 12% compounded monthly. How long will it take him to pay off the purchase if he makes payments of $30 a month?Try it Now Answers
 This is a payout annuity problem. She can pull out $1833.60 a quarter.
 This is a savings annuity problem. He will have saved up $7,524.11/
 This is compound interest problem. She would need to deposit $22,386.46.
 This is a loans problem. She can buy $4,609.33 of new equipment.
 This is a savings annuity problem. You would need to save $200.46 each month
 Solving for
N gives 3.396. It will take about 3.4 years to pay off the purchase.
Monthly payment  $\displaystyle{d}={30}$ 
12% annual rate  $\displaystyle{r}={0.12}$ 
12 months in 1 year  $\displaystyle{k}={12}$ 
The starting loan  $\displaystyle{P}_{{0}}={1000}$ 
$\displaystyle{1000}=\frac{{{30}{\left({1}{\left({1}+\frac{{0.12}}{{12}}\right)}^{{{N}{\left({12}\right)}}}\right)}}}{{\frac{{0.12}}{{12}}}}$ 
David Lippman, Math in Society, "Finance," licensed under a CC BYSA 3.0 license.