9.9 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?

# Ch 9.9 - solving for TIME. Each of the three formulas has the unknown stuck in
# an exponent, so each one ends with a logarithm.
from math import log10

K = 12

def years_to_grow(P0, goal, r, k=K):            # compound interest
    return log10(goal / P0) / (k * log10(1 + r / k))

def years_to_save(d, goal, r, k=K):             # savings annuity
    return log10(goal * (r / k) / d + 1) / (k * log10(1 + r / k))

def years_to_repay(P0, d, r, k=K):              # loan or credit card
    i = r / k
    if d <= P0 * i:
        return None                             # the payment never touches the balance
    return -log10(1 - P0 * i / d) / (k * log10(1 + i))

P0, APR, GOAL = 2000, 0.06, 4000    # TRY IT: GOAL = 8000 (quadruple)
n = years_to_grow(P0, GOAL, APR)
print(f"1) ${P0:,} at {100 * APR:g}% compounded monthly reaches ${GOAL:,}")
print(f"   {GOAL} = {P0}(1 + {APR}/{K})^({K}N)   ->   {GOAL / P0:g} = {1 + APR / K:.6g}^({K}N)")
print(f"   N = log({GOAL / P0:g}) / ({K} log({1 + APR / K:.6g})) = {n:.3f} years")
print(f"   and quadrupling takes {years_to_grow(P0, 4 * P0, APR):.3f} years, "
      "exactly twice as long - doublings stack\n")

D, S_APR, S_GOAL = 100, 0.03, 10000
print(f"2) saving ${D}/month at {100 * S_APR:g}% reaches ${S_GOAL:,} in "
      f"{years_to_save(D, S_GOAL, S_APR):.3f} years")
for goal in (5000, 10000, 20000, 40000):
    print(f"   ${goal:>6,} takes {years_to_save(D, goal, S_APR):>6.2f} years")
print("   Doubling the goal takes LESS than twice as long: the interest is")
print("   compounding the whole time you save.\n")

LOAN, C_APR = 1000, 0.12
print(f"3) a ${LOAN:,} laptop on a credit card at {100 * C_APR:g}%:")
print("   monthly payment   time to clear it   total paid   total interest")
for pay in (10, 15, 20, 25, 30, 50, 100):
    t = years_to_repay(LOAN, pay, C_APR)
    if t is None:
        owed = LOAN * C_APR / K
        print(f"   ${pay:>14}   never - the interest alone is ${owed:,.2f} a month")
    else:
        total = pay * K * t
        print(f"   ${pay:>14}   {t:>13.2f} yr   {total:>10,.2f}   {total - LOAN:>13,.2f}")
print(f"\n   The book's case, $30 a month, takes "
      f"{years_to_repay(LOAN, 30, C_APR):.3f} years.")
print("   Look at the $10 row. Below the interest line, time to repay is INFINITE.")

Show solution

This is a compound interest problem, since we are depositing money once and allowing it to grow. In this problem,

$\begin{array}{ll} P_0 = \$2000 & \text{the initial deposit} \\ r = 0.06 & 6\% \text{ annual rate} \\ k = 12 & \text{12 months in 1 year} \end{array}$

So our general equation is $P_{N}=2000\left(1+\frac{0.06}{12}\right)^{N \times 12}$ . 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.

$\begin{array}{ll} 4000=2000\left(1+\frac{0.06}{12}\right)^{N \times 12} & \text{Divide both sides by 2000} \\ 2=(1.005)^{12 N} & \text{To solve for the exponent, take the log of both sides} \\ \log (2)=\log \left((1.005)^{12 N}\right) & \text{Use the exponent property of logs on the right side} \\ \log (2)=12 N \log (1.005) & \text{Now we can divide both sides by } 12\log{1.005} \\ \frac{\log (2)}{12 \log (1.005)}=N & \text{Approximating this to a decimal} \\ N=11.581 & \end{array}$

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 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?

Show solution

This is a savings annuity problem since we are making regular deposits into the account.

$\begin{array}{ll} d = \$100 & \text{the monthly deposit} \\ r = 0.03 & 3\% \text{ annual rate} \\ k = 12 & \text{since we’re doing monthly deposits, we’ll compound monthly} \end{array}$

We don't know $N,$ but we want $P_{N}$ to be $\$ 10,000$

Putting this into the equation:

$\begin{array}{ll} 10,000=\frac{100\left(\left(1+\frac{0.03}{12}\right)^{N(12)}-1\right)}{\left(\frac{0.03}{12}\right)} & \text{ Simplifying the fractions a bit} \\ 10,000=\frac{100\left((1.0025)^{12 N}-1\right)}{0.0025} & \end{array}$

We want to isolate the exponential term, $1.0025^{12 N}$ , so multiply both sides by 0.0025

$\begin{array}{ll} 25=100\left((1.0025)^{12 N}-1\right) & \text{Divide both sides by 100} \\ 0.25=(1.0025)^{12 N}-1 & \text{Add 1 to both sides} \\ 1.25=(1.0025)^{12 N} & \text{Now take the log of both sides} \\ \log (1.25)=\log \left((1.0025)^{12 \mathrm{N}}\right) & \text{Use the exponent property of logs} \\ \log (1.25)=12 N \log (1.0025) & \text{Divide by 12log(1.0025)} \\ \frac{\log (1.25)}{12 \log (1.0025)}=N & \text{Approximating to a decimal} \\ N=7.447\text{ years} & \end{array}$

It will take about 7.447 years to grow the account to $\$ 10,000$ .

Your Turn

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?

Answer

$\begin{array}{ll} d= \$30 & \text{The monthly payments} \\ r = 0.12 & 12\% \text{ annual rate} \\ k = 12 & \text{since we’re doing monthly deposits} \\ P_0 = \$1000 & \text{we’re starting with a \$1,000 loan} \end{array}$

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.

Adapted from Math in Society by David Lippman, hosted on LibreTexts (math.libretexts.org) and licensed under CC BY-SA 3.0. Changes were made. License: CC-BY-SA-3.0.