7.7: Solving For Time
- Page ID
- 69498
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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.
This section assumes you’ve covered solving exponential equations using logarithms, either in prior classes or in the growth models chapter.
If you invest $2,000 at 6% compounded monthly, how long will it take the account to double in value?
Solution
This is a compound interest problem since we are depositing money once and allowing it to grow. In this problem,
The initial deposit:
\(P_0 = $2000\)
6% annual rate:
\(r = 0.06\)
12 months in 1 year:
\(k = 12\)
So, our general equation is \(P_N = 2000 \left( 1 + \dfrac{0.06}{12} \right)^{N \times 12} \). We also know that we want our ending amount to be double of $2,000, which is $4,000, so we’re looking for \(N\) so that \(P_N = 4,000\). To solve this, we set our equation for \(P_N\) equal to \(4,000\).
Divide both sides by 2,000.
\(4000 = 2000 \left( 1 + \dfrac{0.06}{12} \right)^{N \times 12} \)
To solve for the exponent, take the log of both sides.
\(2 = (1.005)^{12N}\)
Use the exponent property of logs on the right side.
\( \log(2) = \log \left( (1.005)^{12N} \right) \)
Now we can divide both sides by \(12 \log(1.005)\)
\( \log(2) = 12N \log (1.005) \)
\(\dfrac{\log(2)}{12 \log(1.005)} = N\)
Approximating this to three decimal places, we obtain \(N = 11.581\). Thus, 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.
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?
Solution
This is a savings annuity problem since we are making regular deposits into the account.
The monthly deposit:
\(d = $100\)
3% annual rate:
\(r = 0.03\)
Since we’re doing monthly deposits, we’ll compound monthly:
\(k = 12\)
We don’t know \(N\), but we want \(P_N\) to be \($10,000\).
Putting this into the equation, we get:
\(10000 = \dfrac{100 \left( \left( 1+ \dfrac{0.03}{12} \right)^{N(12)} -1 \right) }{\left(\dfrac{0.03}{12}\right)} \)
Simplifying the fractions a bit:
\(10000 = \dfrac{100 \left( \left( 1.0025 \right)^{12N} -1 \right) }{\left( 0.0025 \right)} \)
We want to isolate the exponential term, \(1.0025^{12N}\), so multiply both sides by \(0.0025\).
Divide both sides by 100:
\(25 = 100 \left( (1.0025)^{12N} - 1 \right) \)
Add 1 to both sides:
\(0.25 = (1.0025)^{12N} - 1 \)
Now take the log of both sides:
\(1.25 = (1.0025)^{12N} \)
Use the exponent property of logs:
\(\log(1.25) = \log \left( (1.0025)^{12N} \right) \)
Divide by \(12 \log(1.0025)\)
\(\log(1.25) = 12N \log \left( 1.0025 \right) \)
\(\dfrac{\log(1.25)}{12N \log \left( 1.0025 \right)} = N \)
Approximating to three decimal places, we get \(N = 7.447\) years. Thus, it will take about \(7.447\) years to grow the account to \($10,000\).
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?
1. \(I\) = \($30\) of interest
\(P_0\) = \($500\) principal
\(r\) = unknown
\(t\) = \(1\) month
Using \(I = P_0rt\), we get \(30 = 500 \cdot r \cdot 1\). Solving, we get \(r = 0.06\), or \(6 \%\). Since the time was monthly, this is the monthly interest. The annual rate would be \(12\) times this: \(72\%\) interest.
2. The daily deposit: \(d = $5\)
3% annual rate: \(r = 0.03\)
Since we’re doing daily deposits, we’ll compound daily: \(k = 365\)
We want the amount after 10 years: \(N = 10\)
\(P_{10} = \dfrac{5 \left( \left( 1 + \dfrac{0.03}{365} \right)^{365 \times 10} -1 \right) }{\dfrac{0.03}{365}} = $21,282.07\)
We would have deposited a total of \($5 \cdot 365 \cdot 10 = $18250\), so \($3,032.07\) is from interest
3. \(d = \text{ unknown}\)
4% annual rate: \(r = 0.04\)
Since we’re doing annual scholarships: \(k = 1\)
20 years: \(N = 20\)
We’re starting with $100,000: \(P_0 = 100,000\)
\(100000 = \dfrac{d \left( 1 - \left(1 + \dfrac{0.04}{1} \right)^{-20 \times 1} \right) }{\dfrac{0.04}{1}}\)
Solving for \(d\) gives \($7,358.18\) each year that they can give in scholarships.
It is worth noting that usually donors instead specify that only interest is to be used for scholarship, which makes the original donation last indefinitely. If this donor had specified that, \($100,000(0.04) = $4,000\) a year would have been available.
4. \(d = \text{ unknown}\)
16% annual rate: \(r = 0.16\)
Since we’re making monthly payments: \(k = 12\)
2 years to repay: \(N = 2\)
We’re starting with a $3,000 loan: \(P_0 = 3,000\)
\(3000 = \dfrac{d \left( 1 - \left(1 + \dfrac{0.16}{12} \right)^{-2 \times 12} \right) }{\dfrac{0.16}{12}}\)
Solving for \(d\) gives \($146.89\) as monthly payments.
In total, she will pay \($3,525.36\) to the store, meaning she will pay \($525.36\) in interest over the two years.
5. The monthly payments: \(d = $30\)
12% annual rate: \(r = 0.12\)
Since we’re making monthly payments: \(k = 12\)
2 years to repay: \(N = 2\)
We’re starting with a $1,000 loan: \(P_0 = 1,000\)
We are solving for \(N\), the time to pay off the loan
\(1000 = \dfrac{30 \left( 1 - \left(1 + \dfrac{0.12}{12} \right)^{-N (12)} \right) }{\dfrac{0.12}{12}}\)
Solving for \(N\) gives \(3.396\). It will take about \(3.4\) years to pay off the purchase.