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WealthWiseGuys Practical Mathematics

Chapter 26

Exponential Growth, Logarithms, and Financial Modeling

Repeated change, target times, and the cost of money

Repeated percent changes power many financial models, but assumptions matter as much as formulas. Use hypothetical business and consumer situations to calculate growth, reverse it with logarithms, and compare borrowing costs without treating forecasts as promises.

26.1

Exponential Growth and Decay

Learning objectives

After this section you will be able to:

  1. Distinguish repeated addition from repeated multiplication.
  2. Calculate growth and decay with a constant percent rate.
  3. Interpret model predictions and their limits.

Real-life situation

Exponential Growth and Decay in practice

A hypothetical Philadelphia repair club tracks a $500 equipment reserve and a $1,000 machine. Rates are modeling assumptions, not promised returns. Round final money amounts to cents; keep intermediate powers unrounded.

PeriodReserve at 10% growthMachine at 20% decay
05001000
1550800
2605640
Figure 26.1 Exponential Growth and Decay (hypothetical data)

The concept

Linear change adds the same amount each period. Exponential change multiplies by the same factor. Adding $50 to $500 produces $550 and then $600; growing by 10% produces $550 and then $605 because the second increase uses a larger base.

A percent rate must become a decimal before entering the model. A 10% increase gives a factor of 1.10, while a 20% decrease gives 0.80. The exponent counts complete periods; match the rate period to the time unit.

For decay, the remaining amount is multiplied again, not reduced by the original dollar loss each time. Under a 20% annual decline, a $1,000 machine loses $200 in year one but $160 in year two. This model estimates resale value; it is not an accounting or tax rule.

A constant-rate model is a conditional forecast. Real sales, equipment values, and account rates change. State the starting value, period, and assumption, and do not round every intermediate result unless the actual account credits rounded amounts each period.

Exponential Growth and Decay: solution map

  1. 1

    Identify the question

    Model $500 growing by 10% per year for 3 years.

  2. 2

    Calculation 1

    Convert 10% to 0.10.

  3. 3

    Calculation 2

    Factor = 1 + 0.10 = 1.10.

  4. 4

    Calculation 3

    Compute 1.10³ = 1.331.

  5. 5

    Calculation 4

    Multiply $500 × 1.331 = $665.50.

  6. 6

    Interpret and check

    The modeled balance is $665.50.

Figure 26.1.V Exponential Growth and Decay: solution map. Hypothetical teaching data.

Money Words

Plain-language words you will use in this lesson — learn them before we reach the formulas.

  • compound interest

    Interest earned on both the original amount and the interest already earned.

  • future value

    What an amount of money today will be worth at a later date with interest.

  • present value

    What a future amount of money is worth today at a given interest rate.

Key terms

Exponential model
A model with a constant multiplicative change per period.
Growth factor
One plus the decimal growth rate.
Decay factor
One minus the decimal decline rate.
Initial value
The quantity at period zero.
Exponent
The number of repeated factor multiplications.

Formulas & rules

  • Growth

    A = P(1 + r)ⁿ

  • Decay

    A = P(1 − d)ⁿ

  • Linear comparison

    A = P + kn

Worked examples

Example 1A growing reserve

Model $500 growing by 10% per year for 3 years.

Solution

  1. 1Convert 10% to 0.10.
  2. 2Factor = 1 + 0.10 = 1.10.
  3. 3Compute 1.10³ = 1.331.
  4. 4Multiply $500 × 1.331 = $665.50.

Answer: The modeled balance is $665.50.

Example 2Equipment resale

Estimate a $1,000 machine after 3 years of 20% annual decay.

Solution

  1. 1Convert 20% to 0.20.
  2. 2Remaining factor = 1 − 0.20 = 0.80.
  3. 3Compute 0.80³ = 0.512.
  4. 4Multiply $1,000 × 0.512 = $512.00.

Answer: Estimated value: $512.00, not a guaranteed sale price.

Practice exercises 26.1

  1. Grow $200 by 5% for 2 years.
  2. Reduce $800 by 25% each year for 2 years.
  3. Compare adding $20 yearly to $200 with growing $200 by 10% yearly, after 2 years.
  4. A subscriber count goes 100, 120, 144. Identify the constant factor and rate.
  5. A model predicts 400 orders after two years. Is that a guarantee?

Recap quiz 26.1

1. Which is a 15% decline factor?
2. What is $100 after two 10% increases?
3. Which sequence is exponential?

Section 26.1 recap

  • Use a multiplicative factor for a constant percent change.
  • Match time units and state assumptions; round money at the end.
Answer key with explanations: Section 26.1

Practice exercises

  1. 200 × 1.05² = 220.50, so the result is $220.50.
  2. Factor = 0.75; 800 × 0.75² = $450.00.
  3. Linear: 200 + 2 × 20 = $240.00. Exponential: 200 × 1.10² = $242.00, $2.00 more.
  4. 120 ÷ 100 = 1.2 and 144 ÷ 120 = 1.2; rate = 1.2 − 1 = 20% per period.
  5. No. Repeated multiplication assumes a stable rate; changing demand can change actual orders. Treat 400 as a conditional prediction.

Recap quiz

  1. 0.85: 1 − 0.15 = 0.85.
  2. $121: 100 × 1.10² = $121.00.
  3. 3, 6, 12: 6 ÷ 3 = 12 ÷ 6 = 2, a constant factor.
26.2

Logarithms and Time to a Target

Learning objectives

After this section you will be able to:

  1. Interpret a logarithm as an exponent.
  2. Solve for time in a positive exponential model.
  3. Round a threshold time to a feasible whole period.

Real-life situation

Logarithms and Time to a Target in practice

A hypothetical student business models a $400 reserve growing at 10% annually until it reaches $600. This is a math scenario, not investment advice; money is rounded to cents and times to two decimals unless a whole period is required.

YearModeled reserve
0$400.00
4$585.64
5$644.20
Figure 26.2 Logarithms and Time to a Target (hypothetical data)

The concept

An exponent answers how much results after a known number of periods. A logarithm reverses that question: what exponent produces a specified amount? For example, log base 2 of 8 equals 3 because 2³ = 8.

To solve A = P × bⁿ, first divide by P. Then take the same logarithm on both sides. The power rule brings the exponent down, giving n = log(A ÷ P) ÷ log(b). Common logarithms and natural logarithms give the same ratio.

The logarithm inputs must be positive, and the base must be positive and different from 1. A zero growth rate gives b = 1, so division by log(1) would divide by zero. With no growth, a larger target is never reached by this model.

A decimal answer describes an interpolated model time. If crediting or checking happens only at year-end, round up for the first year at or above a growing target and verify the surrounding years. Decay thresholds also require checking the stated inequality; never round a deadline down automatically.

Logarithms and Time to a Target: solution map

  1. 1

    Identify the question

    When does $400 at modeled 10% annual growth first reach $600 at a year-end?

  2. 2

    Calculation 1

    Divide: 600 ÷ 400 = 1.5.

  3. 3

    Calculation 2

    Solve n = ln(1.5) ÷ ln(1.10) ≈ 4.25 years.

  4. 4

    Calculation 3

    Round up to year 5 for year-end checks.

  5. 5

    Calculation 4

    Verify: year 4 is $585.64; year 5 is $644.20.

  6. 6

    Interpret and check

    Year 5 is the first qualifying year-end.

Figure 26.2.V Logarithms and Time to a Target: solution map. Hypothetical teaching data.

Money Words

Plain-language words you will use in this lesson — learn them before we reach the formulas.

  • inflation

    A general rise in prices over time, which makes money buy less.

  • investment

    Money put into something with the hope it will grow or earn income later.

  • annuity

    A series of equal payments made at regular intervals, such as monthly.

Key terms

Logarithm
The exponent needed to obtain a positive number from a valid base.
Natural logarithm
A logarithm with base e, available as ln on calculators.
Threshold
A target level that must be reached or crossed.
Discrete period
A separate counting interval such as a full year.
Power rule
The rule log(bⁿ) = n × log(b) for positive b.

Formulas & rules

  • Inverse relationship

    log base b of x = n means bⁿ = x

  • Time to target

    n = ln(A ÷ P) ÷ ln(b)

  • Doubling time

    n = ln(2) ÷ ln(1 + r)

Worked examples

Example 1Reserve target

When does $400 at modeled 10% annual growth first reach $600 at a year-end?

Solution

  1. 1Divide: 600 ÷ 400 = 1.5.
  2. 2Solve n = ln(1.5) ÷ ln(1.10) ≈ 4.25 years.
  3. 3Round up to year 5 for year-end checks.
  4. 4Verify: year 4 is $585.64; year 5 is $644.20.

Answer: Year 5 is the first qualifying year-end.

Example 2A shrinking inventory

A supply of 1,600 units halves weekly. When will 200 units remain?

Solution

  1. 1Set 200 = 1,600 × 0.5ⁿ.
  2. 2Divide: 200 ÷ 1,600 = 0.125.
  3. 3Compute n = ln(0.125) ÷ ln(0.5) = 3.
  4. 4Check: 1,600 × 0.5³ = 200.

Answer: After 3 weeks.

Practice exercises 26.2

  1. Find log base 3 of 81.
  2. A count grows from 50 to 400 by doubling each week. Find the time.
  3. Find doubling time at 5% growth per year, to two decimals.
  4. A $1,000 reserve grows 20% annually. First year-end at or above $1,700?
  5. Can $300 at 0% growth reach $450 without deposits?

Recap quiz 26.2

1. log base 10 of 1,000 equals what?
2. A target is crossed at model time 6.2 years. First qualifying year-end?
3. Which logarithm input is valid in real-number math?

Section 26.2 recap

  • Logarithms solve for unknown exponents.
  • Verify whole-period thresholds in the original model.
Answer key with explanations: Section 26.2

Practice exercises

  1. 3⁴ = 81, so the logarithm is 4.
  2. 400 ÷ 50 = 8 = 2³, so 3 weeks.
  3. ln(2) ÷ ln(1.05) ≈ 14.21 years in the interpolated model.
  4. ln(1.7) ÷ ln(1.2) ≈ 2.91. Year 2: $1,440.00; year 3: $1,728.00. Choose year 3.
  5. No: 300 × 1ⁿ = 300 in every period. The logarithm time formula has denominator ln(1) = 0 and cannot be used.

Recap quiz

  1. 3: 10³ = 1,000.
  2. 7: Year 6 is too early; use year 7 and verify.
  3. 0.5: 0.5 is positive; real logarithms require positive inputs.
26.3

Compounding, Loans, and Fair Comparisons

Learning objectives

After this section you will be able to:

  1. Calculate compound balances with different crediting periods.
  2. Calculate a fixed-payment loan and total borrowing cost.
  3. Compare alternatives using consistent assumptions.

Real-life situation

Compounding, Loans, and Fair Comparisons in practice

A hypothetical Philadelphia student compares a savings model and a fixed-payment equipment loan. All quoted rates are invented, not offers. Ignore fees and taxes unless stated; use unrounded formula values and round money to cents at the end.

ModelPrincipalAnnual rateSchedule
Savings$1,00012% nominalMonthly compounding
Loan$1,00012% nominal12 monthly payments
Figure 26.3 Compounding, Loans, and Fair Comparisons (hypothetical data)

The concept

A nominal annual rate does not by itself tell you how often interest is credited. Divide that rate by the number of periods per year, then multiply the years by the same number. Monthly compounding at a nominal 12% uses 1% per month, not 12% per month.

The effective annual rate describes the actual one-year growth factor minus 1 under the stated compounding rule. It allows a cleaner comparison of savings models with different crediting schedules, assuming no fees or withdrawals. It is not a forecast of variable investment returns.

A fixed-payment installment loan charges interest on the unpaid balance. Each payment covers that period's interest first, then reduces principal. For a positive periodic rate, the payment formula balances the present value of all end-of-period payments against the borrowed amount.

Total scheduled payments minus principal gives borrowing cost when there are no fees. A rounded scheduled payment may require an adjusted final payment, so formula totals are estimates until an actual repayment schedule is supplied. Compare total cash outflow, term, fees, and ownership consequences, not just the smallest monthly payment.

Compounding, Loans, and Fair Comparisons: solution map

  1. 1

    Identify the question

    Find $1,000 after one year at 12% nominal, compounded monthly.

  2. 2

    Calculation 1

    Monthly rate = 0.12 ÷ 12 = 0.01.

  3. 3

    Calculation 2

    Number of periods = 12 × 1 = 12.

  4. 4

    Calculation 3

    A = 1,000 × 1.01¹² ≈ $1,126.83.

  5. 5

    Calculation 4

    EAR = (1.01¹² − 1) × 100% ≈ 12.68%.

  6. 6

    Interpret and check

    Balance $1,126.83; effective annual rate about 12.68%.

Figure 26.3.V Compounding, Loans, and Fair Comparisons: solution map. Hypothetical teaching data.

Money Words

Plain-language words you will use in this lesson — learn them before we reach the formulas.

  • yield

    The income an investment pays, shown as a percent of its price.

  • return

    The money or gain an investment produces, often shown as a percent.

  • risk

    The chance that an investment or decision will turn out worse than expected.

Key terms

Nominal annual rate
An annual quoted rate before the effect of within-year compounding.
Effective annual rate
The equivalent one-year growth rate after compounding.
Principal
The original borrowed or deposited amount, or unpaid loan balance in context.
Amortization
Reduction of a loan balance through scheduled payments.
Periodic rate
The rate applicable to one payment or crediting period.

Formulas & rules

  • Compound balance

    A = P(1 + r ÷ m)^(mt)

  • Effective annual rate

    EAR = (1 + r ÷ m)^m − 1

  • Loan payment

    M = P × i ÷ [1 − (1 + i)^(−n)]

  • Borrowing cost without fees

    Cost = total payments − P

Worked examples

Example 1Monthly compounding

Find $1,000 after one year at 12% nominal, compounded monthly.

Solution

  1. 1Monthly rate = 0.12 ÷ 12 = 0.01.
  2. 2Number of periods = 12 × 1 = 12.
  3. 3A = 1,000 × 1.01¹² ≈ $1,126.83.
  4. 4EAR = (1.01¹² − 1) × 100% ≈ 12.68%.

Answer: Balance $1,126.83; effective annual rate about 12.68%.

Example 2Equipment loan

Estimate payments and interest on $1,000 at 12% nominal, paid monthly over one year, with no fees.

Solution

  1. 1i = 0.12 ÷ 12 = 0.01; n = 12.
  2. 2M = 1,000 × 0.01 ÷ [1 − 1.01^(−12)] ≈ $88.85.
  3. 3Using the unrounded formula payment, total ≈ $1,066.19.
  4. 4Subtract principal: $1,066.19 − $1,000.00 = $66.19.

Answer: About $88.85 monthly and $66.19 interest; the final payment may be adjusted for rounding.

Practice exercises 26.3

  1. Find $500 after 2 years at 4% compounded annually.
  2. Find the effective annual rate for 8% nominal compounded semiannually.
  3. A $600 zero-interest loan has 12 equal monthly payments and no fees. Find each payment.
  4. A loan advances $900 and requires 10 payments of $96, with no other fees. Find borrowing cost.
  5. On a loan balance of $1,000, the monthly rate is 1% and payment is $90. Find first-month interest and ending balance.

Recap quiz 26.3

1. 12% nominal compounded monthly means what monthly rate?
2. A no-fee $500 loan is repaid with $560 total. Cost?
3. What should accompany a monthly-payment comparison?

Section 26.3 recap

  • Match the periodic rate to the number of compounding or payment periods.
  • Use total costs and stated assumptions, not a monthly payment alone.
Answer key with explanations: Section 26.3

Practice exercises

  1. 500 × 1.04² = $540.80.
  2. Periodic rate = 0.08 ÷ 2 = 0.04; 1.04² − 1 = 0.0816 = 8.16%.
  3. At zero interest use principal ÷ count, not the positive-rate formula: 600 ÷ 12 = $50.00.
  4. Total = 10 × 96 = $960.00; cost = 960 − 900 = $60.00.
  5. Interest = 1,000 × 0.01 = $10.00. Principal repaid = 90 − 10 = $80.00. Ending balance = $920.00.

Recap quiz

  1. 1%: 12% ÷ 12 = 1%.
  2. $60: 560 − 500 = $60.00.
  3. Term, fees, and total cash paid: A lower payment can hide a longer term and higher total cost.

Chapter 26 Organizer and Review

Chapter Organizer

Every section at a glance: key formulas, key terms, and the main idea.

  • 26.1 Exponential Growth and Decay

    Key formulas

    • Growth: A = P(1 + r)ⁿ
    • Decay: A = P(1 − d)ⁿ
    • Linear comparison: A = P + kn

    Key terms

    Exponential model, Growth factor, Decay factor, Initial value, Exponent

    Remember

    Use a multiplicative factor for a constant percent change.

  • 26.2 Logarithms and Time to a Target

    Key formulas

    • Inverse relationship: log base b of x = n means bⁿ = x
    • Time to target: n = ln(A ÷ P) ÷ ln(b)
    • Doubling time: n = ln(2) ÷ ln(1 + r)

    Key terms

    Logarithm, Natural logarithm, Threshold, Discrete period, Power rule

    Remember

    Logarithms solve for unknown exponents.

  • 26.3 Compounding, Loans, and Fair Comparisons

    Key formulas

    • Compound balance: A = P(1 + r ÷ m)^(mt)
    • Effective annual rate: EAR = (1 + r ÷ m)^m − 1
    • Loan payment: M = P × i ÷ [1 − (1 + i)^(−n)]
    • Borrowing cost without fees: Cost = total payments − P

    Key terms

    Nominal annual rate, Effective annual rate, Principal, Amortization, Periodic rate

    Remember

    Match the periodic rate to the number of compounding or payment periods.

Cumulative Review

Mixed practice from this chapter and the two chapters before it.

  1. Section 26.1

    A model predicts 400 orders after two years. Is that a guarantee?

    Show answer

    No. Repeated multiplication assumes a stable rate; changing demand can change actual orders. Treat 400 as a conditional prediction.

  2. Section 26.2

    Can $300 at 0% growth reach $450 without deposits?

    Show answer

    No: 300 × 1ⁿ = 300 in every period. The logarithm time formula has denominator ln(1) = 0 and cannot be used.

  3. Section 26.3

    On a loan balance of $1,000, the monthly rate is 1% and payment is $90. Find first-month interest and ending balance.

    Show answer

    Interest = 1,000 × 0.01 = $10.00. Principal repaid = 90 − 10 = $80.00. Ending balance = $920.00.

  4. Review of Section 24.3

    Is (3, 5) feasible for the shop?

    Show answer

    Total packs: 3 + 5 = 8 ≤ 8. Machine time: 2 × 3 + 5 = 11 ≤ 12. Both counts are nonnegative integers, so yes.

  5. Review of Section 25.3

    Find the vertex and maximum of P(q) = −q² + 12q − 20 for q ≥ 0.

    Show answer

    qᵥ = −12 ÷ (−2) = 6. P(6) = −36 + 72 − 20 = 16. Since a < 0, the maximum is 16 at q = 6.

All prices, stores, people, and accounts in this chapter are hypothetical and for teaching only. Product references are generic categories; no brand endorses or is endorsed by this textbook.

Chapter 25Chapter 27