$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 7.8: Solving for Time

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

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 15

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, P0 =$2000                  the initial deposit

r = 0.06                       6% annual rate

k = 12                          12 months in 1 year

So our general equation is

ParseError: EOF expected (click for details)
Callstack:
at (Courses/Lumen_Learning/Book:_Mathematics_for_the_Liberal_Arts_(Lumen)/07:_Finance/07.8:_Solving_for_Time), /content/body/div[1]/div[1]/p[6]/img/@alt, line 1, column 4

. 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 PN = 4000. To solve this, we set our equation for PN equal to 4000.

$4000=2000{{\left(1+\frac{0.06}{12}\right)}^{N\times12}}$              Divide both sides by 2000

$2={{\left(1.005\right)}^{12N}}$                                   To solve for the exponent, take the log of both sides

$\log\left(2\right)=\log\left({{\left(1.005\right)}^{12N}}\right)$             Use the exponent property of logs on the right side

$\log\left(2\right)=12N\log\left(1.005\right)$                     Now we can divide both sides by 12log(1.005)

$\frac{\log\left(2\right)}{12\log\left(1.005\right)}=N$                              Approximating this to a decimal

N = 11.581

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.

### Example 16

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.

d = $100 the monthly deposit r = 0.03 3% annual rate k = 12 since we’re doing monthly deposits, we’ll compound monthly We don’t know N, but we want PN to be$10,000.

Putting this into the equation:

$10,000=\frac{100\left({{\left(1+\frac{0.03}{12}\right)}^{N(12)}}-1\right)}{\left(\frac{0.03}{12}\right)}$                        Simplifying the fractions a bit

$10,000=\frac{100\left({{\left(1.0025\right)}^{12N}}-1\right)}{0.0025}$

We want to isolate the exponential term, 1.002512N, so multiply both sides by 0.0025

$25=100\left({{\left(1.0025\right)}^{12N}}-1\right)$                             Divide both sides by 100

$0.25={{\left(1.0025\right)}^{12N}}-1$                                     Add 1 to both sides

$1.25={{\left(1.0025\right)}^{12N}}$                                        Now take the log of both sides

$\log\left(1.25\right)=\log\left({{\left(1.0025\right)}^{12N}}\right)$                              Use the exponent property of logs

$\log\left(1.25\right)=12N\log\left(1.0025\right)$                          Divide by 12log(1.0025)

$\frac{\log\left(1.25\right)}{12\log\left(1.0025\right)}=N$                                               Approximating to a decimal

N = 7.447 years

It 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 1. I =$30 of interest

P0 = $500 principal r = unknown t = 1 month Using I = P0rt, we get 30 = 500·r·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. d =$5              the daily deposit

r = 0.03           3% annual rate

k = 365            since we’re doing daily deposits, we’ll compound daily

N = 10                         we want the amount after 10 years

${{P}_{10}}=\frac{5\left({{\left(1+\frac{0.03}{365}\right)}^{365\times10}}-1\right)}{\frac{0.03}{365}}=$$21,282.07 We would have deposited a total of$5·365·10 = $18,250, so$3,032.07 is from interest

3.

d = unknown

r = 0.04                       4% annual rate

k = 1                           since we’re doing annual scholarships

N = 20                                     20 years

P0 = 100,000               we’re starting with $100,000 $100,000=\frac{d\left(1-{{\left(1+\frac{0.04}{1}\right)}^{-20\times1}}\right)}{\frac{0.04}{1}}$ 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.

### Try it Now Answers continued

4.

d = unknown

r = 0.16                       16% annual rate

k = 12                         since we’re making monthly payments

N = 2                           2 years to repay

P0 = 3,000                   we’re starting with a $3,000 loan $3,000=\frac{d\left(1-{{\left(1+\frac{0.16}{12}\right)}^{-2\times12}}\right)}{\frac{0.16}{12}}$ 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.

6.

d = $30 The monthly payments r = 0.12 12% annual rate k = 12 since we’re making monthly payments P0 = 1,000 we’re starting with a$1,000 loan

We are solving for N, the time to pay off the loan

$1,000=\frac{30\left(1-{{\left(1+\frac{0.12}{12}\right)}^{-N(12)}}\right)}{\frac{0.12}{12}}$

Solving for N gives 3.396. It will take about 3.4 years to pay off the purchase.