Tasks 01-05 - Logarithms & Substitution

Advanced Algebraic Tools

A: Skills

Problem 1: Substitution for Factorization (xx)

Factor the following expressions completely using substitution:

  1. \(x^4 + 5x^2 + 6\)

  2. \(4x^4 - 13x^2 + 9\)

  3. \(x^6 - 9x^3 + 8\)

  4. \((x^2 + 3x)^2 - 2(x^2 + 3x) - 24\)

  5. \(16x^4 - 8x^2 + 1\)

  1. \(x^4 + 5x^2 + 6\)
    • Let \(u = x^2\): \(u^2 + 5u + 6 = (u + 2)(u + 3)\)
    • Substitute back: \((x^2 + 2)(x^2 + 3)\)
  2. \(4x^4 - 13x^2 + 9\)
    • Let \(u = x^2\): \(4u^2 - 13u + 9 = (4u - 9)(u - 1)\)
    • Substitute back: \((4x^2 - 9)(x^2 - 1) = (2x + 3)(2x - 3)(x + 1)(x - 1)\)
  3. \(x^6 - 9x^3 + 8\)
    • Let \(u = x^3\): \(u^2 - 9u + 8 = (u - 1)(u - 8)\)
    • Substitute back: \((x^3 - 1)(x^3 - 8)\)
    • Factor further: \((x - 1)(x^2 + x + 1)(x - 2)(x^2 + 2x + 4)\)
  4. \((x^2 + 3x)^2 - 2(x^2 + 3x) - 24\)
    • Let \(u = x^2 + 3x\): \(u^2 - 2u - 24 = (u - 6)(u + 4)\)
    • Substitute back: \((x^2 + 3x - 6)(x^2 + 3x + 4)\)
  5. \(16x^4 - 8x^2 + 1\)
    • Let \(u = x^2\): \(16u^2 - 8u + 1 = (4u - 1)^2\)
    • Substitute back: \((4x^2 - 1)^2 = [(2x + 1)(2x - 1)]^2 = (2x + 1)^2(2x - 1)^2\)

Problem 2: Logarithm Fundamentals (x)

Evaluate the following without a calculator:

  1. \(\log_2(64)\)

  2. \(\log_3(\frac{1}{27})\)

  3. \(\log_5(125)\)

  4. \(\log_{10}(0.001)\)

  5. \(\log_4(8)\) (Hint: Express 8 as a power of 2, and 4 as a power of 2)

  6. \(\ln(e^3)\)

  7. \(\frac{\log(81)}{\log(3)}\)

  8. \(\frac{\ln(32)}{\ln(4)}\)

Hint for g) and h): read the change of base formula from right to left.

  1. \(\log_2(64) = \log_2(2^6) = 6\)

  2. \(\log_3(\frac{1}{27}) = \log_3(3^{-3}) = -3\)

  3. \(\log_5(125) = \log_5(5^3) = 3\)

  4. \(\log_{10}(0.001) = \log_{10}(10^{-3}) = -3\)

  5. \(\log_4(8)\): Since \(4 = 2^2\) and \(8 = 2^3\)

    • Let \(\log_4(8) = x\), then \(4^x = 8\)
    • \((2^2)^x = 2^3\)
    • \(2^{2x} = 2^3\)
    • \(2x = 3\), so \(x = \frac{3}{2}\)

  6. \(\ln(e^3) = 3\)

  7. \(\frac{\log(81)}{\log(3)} = \log_3(81) = \log_3(3^4) = 4\) (change of base formula)

  8. \(\frac{\ln(32)}{\ln(4)} = \log_4(32)\): Since \(4 = 2^2\) and \(32 = 2^5\)

    • Let \(\log_4(32) = x\), then \(4^x = 32\)
    • \((2^2)^x = 2^5\)
    • \(2^{2x} = 2^5\)
    • \(2x = 5\), so \(x = \frac{5}{2}\)

Problem 3: Logarithm Laws (x)

Simplify the following expressions using logarithm laws:

  1. \(\log_2(16) + \log_2(8) - \log_2(4)\)

  2. \(\log(50) + \log(20) - \log(100)\)

  3. \(3\log_5(x) - \log_5(x^2) + \log_5(25)\)

  4. \(\log_3(81) - 2\log_3(9) + \log_3(27)\)

  5. Express as a single logarithm: \(\frac{1}{2}\ln(x) + 3\ln(y) - \ln(z)\)

  1. \(\log_2(16) + \log_2(8) - \log_2(4)\)
    • Direct: \(4 + 3 - 2 = 5\)
    • Or: \(\log_2(\frac{16 \times 8}{4}) = \log_2(32) = 5\)
  2. \(\log(50) + \log(20) - \log(100)\)
    • \(= \log(\frac{50 \times 20}{100}) = \log(10) = 1\)
  3. \(3\log_5(x) - \log_5(x^2) + \log_5(25)\)
    • \(= \log_5(x^3) - \log_5(x^2) + \log_5(25)\)
    • \(= \log_5(\frac{x^3 \times 25}{x^2}) = \log_5(25x) = \log_5(25) + \log_5(x)\)
    • \(= 2 + \log_5(x)\)
  4. \(\log_3(81) - 2\log_3(9) + \log_3(27)\)
    • \(= 4 - 2(2) + 3 = 4 - 4 + 3 = 3\)
  5. \(\frac{1}{2}\ln(x) + 3\ln(y) - \ln(z)\)
    • \(= \ln(x^{1/2}) + \ln(y^3) - \ln(z)\)
    • \(= \ln(\frac{\sqrt{x} \cdot y^3}{z})\)

Problem 4: Solving Logarithmic and Exponential Equations (xx)

Solve the following equations:

  1. \(\log_3(x + 5) = 2\)

  2. \(\log(2x - 1) - \log(x + 2) = 0\)

  3. \(2^{\log_2(x)} = 8\)

  4. \(\log_x(49) = 2\)

  5. \(7^{2x-3} = 343\)

  6. \(6^x = 50\) (exactly and rounded to three decimal places)

  7. \(9^x - 12 \cdot 3^x + 27 = 0\)

Hint for d): the base of a logarithm must be positive and different from 1.

Hint for g): \(9^x = (3^x)^2\). Substitute \(u = 3^x\).

  1. \(\log_3(x + 5) = 2\)
    • \(x + 5 = 3^2 = 9\)
    • \(x = 4\)
  2. \(\log(2x - 1) - \log(x + 2) = 0\)
    • \(\log(\frac{2x - 1}{x + 2}) = 0\)
    • \(\frac{2x - 1}{x + 2} = 10^0 = 1\)
    • \(2x - 1 = x + 2\)
    • \(x = 3\)
  3. \(2^{\log_2(x)} = 8\)
    • By definition: \(2^{\log_2(x)} = x\)
    • So \(x = 8\)
  4. \(\log_x(49) = 2\)
    • \(x^2 = 49\)
    • \(x = 7\) (x must be positive and ≠ 1)
  5. \(7^{2x-3} = 343\)
    • \(343 = 7^3\), so \(7^{2x-3} = 7^3\)
    • \(2x - 3 = 3\)
    • \(x = 3\)
  6. \(6^x = 50\)
    • Take logarithms: \(\ln(6^x) = \ln(50)\)
    • Power rule: \(x \cdot \ln(6) = \ln(50)\)
    • \(x = \frac{\ln(50)}{\ln(6)} = \log_6(50) \approx 2.183\)
    • Check: \(6^{2.183} \approx 50.0\)
  7. \(9^x - 12 \cdot 3^x + 27 = 0\)
    • Let \(u = 3^x\): \(u^2 - 12u + 27 = (u - 3)(u - 9) = 0\)
    • A product is zero if one of its factors is zero: \(u = 3\) or \(u = 9\)
    • Substitute back: \(3^x = 3 \Rightarrow x = 1\), and \(3^x = 9 = 3^2 \Rightarrow x = 2\)

Problem 5: Binomial (xx)

  1. Expand completely: \((2x - 3)^4\)

  2. Expand completely: \((3x + 1)^6\)

Hint for b): use row 6 of Pascal’s triangle.

  1. \((2x - 3)^4\)
    • Coefficients: 1, 4, 6, 4, 1
    • \(= (2x)^4 + 4(2x)^3(-3) + 6(2x)^2(-3)^2 + 4(2x)(-3)^3 + (-3)^4\)
    • \(= 16x^4 - 96x^3 + 216x^2 - 216x + 81\)
  2. \((3x + 1)^6\)
    • Coefficients (row 6): 1, 6, 15, 20, 15, 6, 1
    • \(= (3x)^6 + 6(3x)^5 \cdot 1 + 15(3x)^4 \cdot 1^2 + 20(3x)^3 \cdot 1^3 + 15(3x)^2 \cdot 1^4 + 6(3x) \cdot 1^5 + 1^6\)
    • \(= 729x^6 + 6 \cdot 243x^5 + 15 \cdot 81x^4 + 20 \cdot 27x^3 + 15 \cdot 9x^2 + 6 \cdot 3x + 1\)
    • \(= 729x^6 + 1458x^5 + 1215x^4 + 540x^3 + 135x^2 + 18x + 1\)

B: Set Up & Check

Problem 6: From Text to Mathematics (xx)

Translate each statement into mathematics. Then answer the check question: it tells you whether your translation works.

  1. “The number of coworking desks in a city doubles every 6 years.”

    Check: Which yearly growth factor (the factor by which the number is multiplied each year) follows from your translation, and does it give 4 times as many desks after 12 years?

  2. “A forklift truck loses 12 % of its value per year.”

    Check: Has the forklift lost 24 % of its value after 2 years?

  3. “A manufacturer recalls a faulty electric kettle. Each week, 45 % of the kettles still with customers are returned. The recall ends once at least 95 % of all kettles have been returned.”

    Check: Does the recall end after 5 weeks?

  4. “The number of cheque payments a bank processes halves every 8 years.”

    Check: Are there no cheque payments left after 16 years?

  5. “The revenue of a tea trading company grows by 5.5 % per year. Its doubling time (the time after which the revenue has doubled) is \(T\) years.”

    Check: Which value does the calculator give for \(T\), and does \(1.055^T\) then give about 2?

  1. With \(N_0\) desks today, after \(t\) years: \(N = N_0 \cdot 2^{t/6}\). Yearly growth factor: \(2^{1/6} = \sqrt[6]{2} \approx 1.122\), i.e. about 12.2 % more desks per year.

    Check: yes, \(\left(2^{1/6}\right)^{12} = 2^{12/6} = 2^2 = 4\).

    Typical mistake: “100 % in 6 years, so about 16.7 % per year”. With this rate, \(\left(\frac{7}{6}\right)^6 \approx 2.52\): the number would more than double. Growth over several years multiplies. It does not add.

  2. With \(V_0\) the purchase price, after \(t\) years: \(V = V_0 \cdot 0.88^t\).

    Check: no, \(0.88^2 = 0.7744\), so the forklift has lost 22.56 % of its value.

    Typical mistake: \(V = V_0 \cdot (1 - 0.12t)\). Each year’s loss of 12 % refers to the value at the start of that year, which gets smaller every year.

  3. Each week, 55 % of the kettles still with customers stay there. After \(n\) weeks, the share still with customers is \(0.55^n\), and the share returned is \(1 - 0.55^n\). The recall ends once

    \[1 - 0.55^n \geq 0.95 \iff 0.55^n \leq 0.05\]

    Check: no. \(0.55^5 \approx 0.0503\), so only about 94.97 % have been returned after 5 weeks. After 6 weeks: \(0.55^6 \approx 0.0277\), i.e. about 97.2 % returned.

    Typical mistake: \(0.45n \geq 0.95\), i.e. “done after 3 weeks”. The 45 % refer to the kettles still with customers, which get fewer every week. So the share still out is multiplied by 0.55 each week.

  4. With \(C_0\) cheque payments today, after \(t\) years: \(C = C_0 \cdot 0.5^{t/8}\).

    Check: no, \(0.5^{16/8} = 0.5^2 = 0.25\). After 16 years, a quarter of today’s cheque payments remain.

    Typical mistake: “half is gone after 8 years, the other half after the next 8 years”. The second halving halves only what is left, so it removes a quarter of today’s payments, not a half.

  5. With \(R_0\) the revenue today: \(R_0 \cdot 1.055^T = 2R_0 \iff 1.055^T = 2 \iff T = \log_{1.055}(2) = \frac{\ln(2)}{\ln(1.055)}\).

    Check: \(T \approx 12.95\) years, and \(1.055^{12.95} \approx 2.00\). Yes.

    Typical mistake: \(\frac{\ln(2)}{\ln(1.055)} = \ln\left(\frac{2}{1.055}\right) \approx 0.64\). A quotient of two logarithms is not the logarithm of a quotient.

Problem 7: One Paragraph, All Conditions (xx)

A regional theatre starts an email newsletter. Today it reaches 6,000 readers, and the marketing team expects the reach to grow by 5 % per month. A sponsor pays a bonus if the newsletter reaches at least 10,000 readers within the first 12 months. The campaign budget lasts for at most 18 months.

List every condition in this paragraph as a mathematical statement, with \(t\) the time in months from today and \(N\) the reach. Do not solve anything.

Check: Suppose the reach passes 10,000 readers only after 14 months. Which of your conditions is violated, and which one is still met?

  1. Start: \(N = 6000\) readers at \(t = 0\)
  2. Growth factor: \(1.05\) per month
  3. Reach after \(t\) months: \(N = 6000 \cdot 1.05^t\)
  4. Bonus target: \(6000 \cdot 1.05^t \geq 10000\)
  5. Bonus deadline: \(t \leq 12\)
  6. Budget: \(0 \leq t \leq 18\)

Check: 14 months violate condition 5, so there is no bonus. Condition 6 still holds: the campaign is still running.

Typical mistake: \(N = 6000 + 0.05 \cdot 6000 \cdot t\). A growth of 5 % per month refers to the current reach, so the reach is multiplied by 1.05 every month.

C: Applied

Problem 8: Reserve Account of a Craft Brewery (xx)

A craft brewery puts 12,000 EUR into a reserve account with continuous compounding (the interest is added continuously instead of once a year). After \(t\) years, the balance is \(12000 \cdot e^{rt}\) EUR, where \(r\) is the yearly interest rate as a decimal. After 5 years, the balance is 15,600 EUR.

  1. Determine \(r\) exactly as a logarithm and give it in percent, rounded to two decimal places.

(for verification: \(r = \frac{\ln(1.3)}{5} \approx 0.0525\))

  1. Show without a calculator that the balance after 10 years is 20,280 EUR. Hint: \(e^{10r} = \left(e^{5r}\right)^2\).

  2. Determine the doubling time (the time after which the balance has doubled) exactly and rounded to 0.1 years.

  3. The brewery needs 20,000 EUR for a new brewing kettle in 9 years. Decide whether the reserve account will be enough by then, and determine after how many years the balance reaches 20,000 EUR, rounded to 0.1 years.

a)

\[12000 \cdot e^{5r} = 15600 \iff e^{5r} = \frac{15600}{12000} = 1.3\]

Take the natural logarithm: \(\ln(e^{x}) = x\).

\[5r = \ln(1.3) \iff r = \frac{\ln(1.3)}{5} \approx \frac{0.2624}{5} \approx 0.0525\]

The yearly interest rate is about 5.25 %.

b) With the power rule for exponents and \(e^{5r} = 1.3\) from a):

\[e^{10r} = \left(e^{5r}\right)^2 = 1.3^2 = 1.69\]

\[12000 \cdot 1.69 = 20280 \text{ EUR} \quad \checkmark\]

c) Doubling means a balance of \(2 \cdot 12000\) EUR:

\[12000 \cdot e^{rt} = 24000 \iff e^{rt} = 2 \iff rt = \ln(2)\]

\[t = \frac{\ln(2)}{r} = \frac{5\ln(2)}{\ln(1.3)} \approx 13.2 \text{ years}\]

d)

\[12000 \cdot e^{rt} = 20000 \iff e^{rt} = \frac{5}{3} \iff t = \frac{\ln\left(\frac{5}{3}\right)}{r} = \frac{5\ln\left(\frac{5}{3}\right)}{\ln(1.3)} \approx 9.7 \text{ years}\]

The balance reaches 20,000 EUR only after about 9.7 years, so the reserve account is not enough after 9 years. Check: after 9 years the balance is \(12000 \cdot e^{9r} \approx 19{,}243\) EUR (with the unrounded \(r\)).

Problem 9: Hailstorm Claims at a Car Insurer (xxx)

After a hailstorm, a car insurer receives 2,000 damage claims. Each week, it settles 30 % of the claims that are still open. The other 70 % stay open.

  1. Explain why \(2000 \cdot 0.7^n\) claims are still open after \(n\) weeks, and compute the number of open claims after 3 weeks.

(for verification: 686 claims)

  1. Determine the minimum number of whole weeks until at least 95 % of the claims are settled. Verify your answer with the week before.

Note: the logarithm of a number between 0 and 1 is negative, e.g. \(\ln(0.7) \approx -0.357\). Taking the logarithm of both sides keeps the inequality sign, because the larger of two positive numbers has the larger logarithm. Dividing by a negative number flips the sign.

  1. The management wants at least 95 % of the claims settled within 6 weeks. Determine which share of the open claims the insurer must then settle each week, in percent and rounded up to a whole percent.

Hint: take the sixth root of both sides. For positive numbers, this keeps the inequality sign.

a) Each week, 70 % of the open claims stay open, so the number of open claims is multiplied by 0.7 every week:

  • after 1 week: \(2000 \cdot 0.7\)
  • after 2 weeks: \(2000 \cdot 0.7 \cdot 0.7 = 2000 \cdot 0.7^2\)
  • after \(n\) weeks: \(2000 \cdot 0.7^n\)

After 3 weeks: \(2000 \cdot 0.7^3 = 2000 \cdot 0.343 = 686\) open claims.

b) At least 95 % settled means at most 5 % still open:

\[2000 \cdot 0.7^n \leq 0.05 \cdot 2000 \iff 0.7^n \leq 0.05\]

Take logarithms (the sign stays) and use the power rule:

\[n \cdot \ln(0.7) \leq \ln(0.05)\]

\(\ln(0.7) < 0\), so dividing by it flips the sign:

\[n \geq \frac{\ln(0.05)}{\ln(0.7)} \approx \frac{-2.996}{-0.357} \approx 8.40\]

Only whole weeks count, so round up: \(n = 9\) weeks.

Check:

  • \(n = 8\): \(0.7^8 \approx 0.0576\), so only about 94.2 % are settled (not enough)
  • \(n = 9\): \(0.7^9 \approx 0.0404\), so about 96.0 % are settled \(\checkmark\)

c) Let \(q\) be the share of open claims that stays open each week. After 6 weeks at most 5 % may be open:

\[q^6 \leq 0.05 \iff q \leq \sqrt[6]{0.05} = 0.05^{1/6} \approx 0.6070\]

The share settled each week is \(1 - q \geq 1 - 0.6070 = 0.3930\), i.e. at least 39.3 %. Rounded up to a whole percent: 40 %.

Check: settling 40 % leaves \(0.6^6 \approx 0.0467 \leq 0.05\) open \(\checkmark\). Settling only 39 % leaves \(0.61^6 \approx 0.0515 > 0.05\) open.

Problem 10: Rent Increases for an Optician (xx)

An optician rents a shop for 2,000 EUR per month. The rental contract raises the monthly rent by the same rate \(r\) every year (\(r\) as a decimal, e.g. \(r = 0.05\) for 5 %).

  1. Show with Pascal’s triangle that after three increases the monthly rent is \(2000 \cdot (1 + 3r + 3r^2 + r^3)\) EUR.

  2. Explain what the term \(3r\) and the terms \(3r^2 + r^3\) stand for.

  3. Let \(r = 0.05\). Compute the monthly rent after three increases. The optician estimated it quickly as \(2000 \cdot (1 + 3 \cdot 0.05)\) EUR. Determine the error of this estimate and explain which terms from a) cause it.

(for verification: 2,315.25 EUR)

  1. A second landlord offers a contract that raises the rent by 10 % every two years. Show with row 4 of Pascal’s triangle that \(1.05^4 = 1.21550625\) and decide which contract gives the lower monthly rent after four years.

a) Each increase multiplies the rent by \((1 + r)\), so after three increases the rent is \(2000 \cdot (1 + r)^3\). Row 3 of Pascal’s triangle is \(1, 3, 3, 1\), with \(a = 1\) and \(b = r\):

\[(1 + r)^3 = 1^3 + 3 \cdot 1^2 \cdot r + 3 \cdot 1 \cdot r^2 + r^3 = 1 + 3r + 3r^2 + r^3\]

\[\text{Rent} = 2000 \cdot (1 + 3r + 3r^2 + r^3) \text{ EUR} \quad \checkmark\]

b) \(2000 \cdot 3r\) is three times the first increase, each calculated on the original rent of 2,000 EUR. \(2000 \cdot (3r^2 + r^3)\) is the increase on earlier increases: from the second year on, the rate also applies to the increases already added.

c)

\[2000 \cdot (1 + 0.15 + 3 \cdot 0.0025 + 0.000125) = 2000 \cdot 1.157625 = 2315.25 \text{ EUR}\]

The estimate gives \(2000 \cdot 1.15 = 2300\) EUR, so it is 15.25 EUR too low. The error is exactly the part from the terms \(3r^2 + r^3\):

\[2000 \cdot (0.0075 + 0.000125) = 2000 \cdot 0.007625 = 15.25 \text{ EUR}\]

d) Row 4 of Pascal’s triangle is \(1, 4, 6, 4, 1\):

\[1.05^4 = (1 + 0.05)^4 = 1 + 4 \cdot 0.05 + 6 \cdot 0.05^2 + 4 \cdot 0.05^3 + 0.05^4\]

\[= 1 + 0.2 + 0.015 + 0.0005 + 0.00000625 = 1.21550625 \quad \checkmark\]

After four years, the first contract has raised the rent four times by 5 %, the second contract twice by 10 %:

  • first contract: \(2000 \cdot 1.21550625 \approx 2431.01\) EUR
  • second contract: \(2000 \cdot 1.1^2 = 2000 \cdot 1.21 = 2420\) EUR

The second contract gives the lower rent, about 11.01 EUR per month less.

D: FSP Problem

Work on this problem as in the exam: write down every step. The points show what each part is worth.

FSP Problem: E-Invoicing at an Office-Supplies Distributor [20 pts. total]

An office-supplies distributor sends invoices to its business customers, either on paper or as e-invoices (electronic invoices sent as a file). In 2022 it sent 64,000 paper invoices, in 2025 only 27,000. The number of paper invoices falls by the same percentage every year. In 2022 the distributor also sent 8,000 e-invoices. Their number grows by 25 % every year.

(\(t\) = time in years since 2022, so \(t = 0\) stands for 2022 and \(t = 3\) for 2025)

Part a)

Determine the yearly growth factor of the paper invoices and give a term for the number \(P\) of paper invoices in year \(t\). [3 pts.]

(for verification: \(P = 64000 \cdot 0.75^t\))

Ansatz: \(P = 64000 \cdot q^t\) with the yearly growth factor \(q\).

Data point 2025 (\(t = 3\)):

\[64000 \cdot q^3 = 27000 \iff q^3 = \frac{27000}{64000} = \frac{27}{64}\]

Third root:

\[q = \sqrt[3]{\frac{27}{64}} = \frac{3}{4} = 0.75\]

\[P = 64000 \cdot 0.75^t \quad \checkmark\]

Grading: 1 Pt. ansatz, 1 Pt. third root, 1 Pt. term.

Part b)

Give the yearly rate by which the number of paper invoices falls, in percent, and compute the number of paper invoices the model forecasts for 2028. [2 pts.]

Rate: \(q = 0.75 = 1 - 0.25\), so the number of paper invoices falls by 25 % per year.

Forecast for 2028 (\(t = 6\)):

\[P = 64000 \cdot 0.75^6 \approx 64000 \cdot 0.1780 \approx 11{,}391 \text{ paper invoices}\]

Grading: 1 Pt. rate, 1 Pt. forecast with \(t = 6\).

Part c)

Determine the halving time of the paper invoices (the time after which their number has halved) exactly as a logarithm and rounded to two decimal places. [3 pts.]

Halving means \(P = \frac{1}{2} \cdot 64000\):

\[64000 \cdot 0.75^t = 32000 \iff 0.75^t = 0.5\]

Take the logarithm of both sides and use the power rule:

\[t \cdot \ln(0.75) = \ln(0.5) \iff t = \frac{\ln(0.5)}{\ln(0.75)} = \log_{0.75}(0.5)\]

\[t \approx \frac{-0.6931}{-0.2877} \approx 2.41 \text{ years}\]

Both logarithms are negative, so \(t\) is positive.

Grading: 1 Pt. equation \(0.75^t = 0.5\), 1 Pt. logarithm and power rule, 1 Pt. exact and rounded result.

Part d)

Give a term for the number \(E\) of e-invoices in year \(t\). Show that the model gives equally many e-invoices and paper invoices at

\[t = \frac{3\ln(2)}{\ln(5) - \ln(3)}\]

and determine the first year in which the distributor sends more e-invoices than paper invoices. [5 pts.]

Term: growth by 25 % means the growth factor \(1.25\):

\[E = 8000 \cdot 1.25^t\]

Set both terms equal:

\[8000 \cdot 1.25^t = 64000 \cdot 0.75^t\]

Divide by \(8000\) and by \(0.75^t\) (which is always positive):

\[\frac{1.25^t}{0.75^t} = 8 \iff \left(\frac{1.25}{0.75}\right)^t = 8 \iff \left(\frac{5}{3}\right)^t = 8\]

Take the logarithm of both sides. Power rule: \(\ln(8) = \ln(2^3) = 3\ln(2)\). Quotient rule: \(\ln\left(\frac{5}{3}\right) = \ln(5) - \ln(3)\):

\[t \cdot \big(\ln(5) - \ln(3)\big) = 3\ln(2) \iff t = \frac{3\ln(2)}{\ln(5) - \ln(3)} \quad \checkmark\]

Numerically: \(t \approx \frac{2.0794}{0.5108} \approx 4.07\).

First year: \(t\) counts whole years, so the first year after \(t \approx 4.07\) is \(t = 5\), i.e. 2027. Check:

  • \(t = 4\) (2026): \(P = 64000 \cdot 0.75^4 = 20{,}250 > E = 8000 \cdot 1.25^4 \approx 19{,}531\)
  • \(t = 5\) (2027): \(P \approx 15{,}188 < E \approx 24{,}414\) \(\checkmark\)

Grading: 1 Pt. term \(E\), 1 Pt. equation and division, 1 Pt. \(\left(\frac{5}{3}\right)^t = 8\), 1 Pt. logarithm laws to the given result, 1 Pt. year 2027 with reason.

Part e)

The distributor will end its contract with a printing service in the first year in which it sends at least 95 % fewer paper invoices than in 2022. Determine this year. [4 pts.]

Condition: at least 95 % fewer means at most 5 % of the 64,000 paper invoices of 2022:

\[64000 \cdot 0.75^t \leq 0.05 \cdot 64000 \iff 0.75^t \leq 0.05\]

Logarithm of both sides (the sign stays) and power rule:

\[t \cdot \ln(0.75) \leq \ln(0.05)\]

\(\ln(0.75) < 0\), so dividing by it flips the sign:

\[t \geq \frac{\ln(0.05)}{\ln(0.75)} \approx \frac{-2.9957}{-0.2877} \approx 10.41\]

Round up to whole years: \(t = 11\), i.e. the year 2033.

Check:

  • \(t = 10\) (2032): \(0.75^{10} \approx 0.0563\), only about 94.4 % fewer (not enough)
  • \(t = 11\) (2033): \(0.75^{11} \approx 0.0422\), about 95.8 % fewer \(\checkmark\)

Grading: 1 Pt. inequality \(0.75^t \leq 0.05\), 1 Pt. logarithm with sign flip, 1 Pt. rounding up to 2033, 1 Pt. check with \(t = 10\) and \(t = 11\).

Part f)

In 2022, one paper invoice cost the distributor 1.50 EUR for paper, printing and postage. By 2025, this cost per invoice had risen by 20 %. Compute the total costs of the paper invoices in 2022 and in 2025 and their percentage change. A manager estimates: “The number fell by 57.8 % and the cost per invoice rose by 20 %, so the total costs fell by 37.8 %.” Explain the error in this estimate. [3 pts.]

2022: \(64000 \cdot 1.50 = 96{,}000\) EUR

2025: cost per invoice \(1.50 \cdot 1.2 = 1.80\) EUR, total \(27000 \cdot 1.80 = 48{,}600\) EUR

Change:

\[\frac{48600 - 96000}{96000} = -0.49375 \approx -49.4\,\%\]

Explanation: total costs = number of invoices × cost per invoice, so the growth factors multiply. They do not add:

\[\frac{27000}{64000} \cdot 1.2 = 0.421875 \cdot 1.2 = 0.50625\]

The costs fell to 50.6 %, i.e. by 49.4 %. The 20 % rise applies only to the 42.2 % of the invoices that are left, not to all 64,000.

Grading: 1 Pt. both total costs, 1 Pt. percentage change, 1 Pt. explanation with multiplied growth factors.

Calculator Practice

Problem 11: Calculator: Fraction Calculations (x)

Use your calculator to evaluate the following expressions. Verify your answers are in simplified form.

  1. \(\frac{3}{4} + \frac{5}{6} - \frac{1}{2}\)

  2. \(\frac{7}{12} \times \frac{8}{21}\)

  3. \(2\frac{1}{3} + 1\frac{3}{4}\)

  4. \(\frac{15}{8} \div \frac{5}{12}\)

  1. \(\frac{3}{4} + \frac{5}{6} - \frac{1}{2}\)
    • Find common denominator: \(\frac{9}{12} + \frac{10}{12} - \frac{6}{12} = \frac{13}{12}\)
    • Answer: \(\frac{13}{12}\) or \(1\frac{1}{12}\)
  2. \(\frac{7}{12} \times \frac{8}{21}\)
    • Multiply and simplify: \(\frac{56}{252} = \frac{2}{9}\)
    • Answer: \(\frac{2}{9}\)
  3. \(2\frac{1}{3} + 1\frac{3}{4}\)
    • Convert to improper: \(\frac{7}{3} + \frac{7}{4} = \frac{28}{12} + \frac{21}{12} = \frac{49}{12}\)
    • Answer: \(\frac{49}{12}\) or \(4\frac{1}{12}\)
  4. \(\frac{15}{8} \div \frac{5}{12}\)
    • Multiply by reciprocal: \(\frac{15}{8} \times \frac{12}{5} = \frac{180}{40} = \frac{9}{2}\)
    • Answer: \(\frac{9}{2}\) or \(4\frac{1}{2}\)

Problem 12: Calculator: Percentage Calculations (x)

Use your calculator’s percentage function to solve:

  1. What is 15% of 240?

  2. Increase 180 by 12%

  3. Decrease 500 by 8%

  4. A product costs 85 EUR after a 15% discount. What was the original price?

  1. 15% of 240
    • Enter: 240 × 15 SHIFT ANS =
    • Answer: 36
  2. Increase 180 by 12%
    • Enter: 180 + 180 × 12 SHIFT ANS =
    • Answer: 201.6
  3. Decrease 500 by 8%
    • Enter: 500 - 500 × 8 SHIFT ANS =
    • Answer: 460
  4. Original price if 85 EUR is 15% off
    • 85 EUR = 85% of original → Original = 85 ÷ 0.85
    • Answer: 100 EUR

Problem 13: Calculator: Prime Factorization (x)

Use the FACT function (SHIFT + FACT) to find:

  1. Prime factorization of 360

  2. Prime factorization of 1260

  3. Use prime factorization to simplify \(\frac{84}{126}\)

  1. 360
    • Enter: 360 = then SHIFT + FACT
    • Answer: \(2^3 \times 3^2 \times 5\)
  2. 1260
    • Enter: 1260 = then SHIFT + FACT
    • Answer: \(2^2 \times 3^2 \times 5 \times 7\)
  3. Simplify \(\frac{84}{126}\)
    • 84 = \(2^2 \times 3 \times 7\)
    • 126 = \(2 \times 3^2 \times 7\)
    • GCD = \(2 \times 3 \times 7 = 42\)
    • Answer: \(\frac{84}{126} = \frac{2}{3}\)

Problem 14: Calculator: GCD and LCM (x)

Use the GCD and LCM functions:

  1. Find GCD(48, 72)

  2. Find LCM(12, 18)

  3. Use GCD to simplify \(\frac{144}{180}\)

  4. Find the smallest number divisible by both 15 and 20

  1. GCD(48, 72)
    • Enter: GCD(48; 72) =
    • Answer: 24
  2. LCM(12, 18)
    • Enter: LCM(12; 18) =
    • Answer: 36
  3. Simplify \(\frac{144}{180}\)
    • GCD(144, 180) = 36
    • \(\frac{144}{180} = \frac{144 \div 36}{180 \div 36} = \frac{4}{5}\)
    • Answer: \(\frac{4}{5}\)
  4. Smallest number divisible by 15 and 20
    • LCM(15, 20) = 60
    • Answer: 60

Problem 15: Calculator: Powers and Roots (x)

Calculate the following using your calculator:

  1. \(3^4 + 2^5 - 5^2\)

  2. \(\sqrt{169} + \sqrt[3]{125}\)

  3. \((2^3)^2 - 4^3\)

  4. \(\sqrt[4]{256} \times \sqrt{49}\)

  1. \(3^4 + 2^5 - 5^2\)
    • \(81 + 32 - 25 = 88\)
    • Answer: 88
  2. \(\sqrt{169} + \sqrt[3]{125}\)
    • \(13 + 5 = 18\)
    • Answer: 18
  3. \((2^3)^2 - 4^3\)
    • \(8^2 - 64 = 64 - 64 = 0\)
    • Answer: 0
  4. \(\sqrt[4]{256} \times \sqrt{49}\)
    • \(4 \times 7 = 28\)
    • Answer: 28

Problem 16: Calculator: Memory and Variables (x)

Follow these steps and record each result:

  1. Store the value 12.5 in variable A. Then calculate \(3 \times A + 7\).

  2. Calculate \(\sqrt{50}\) and store the result in B. Then find \(B^2\).

  3. If A = 5 and B = 3, use stored variables to calculate \(A^2 - B^2\).

  1. Store 12.5 in A, calculate \(3 \times A + 7\)
    • \(3 \times 12.5 + 7 = 37.5 + 7 = 44.5\)
    • Answer: 44.5
  2. Store \(\sqrt{50}\) in B, find \(B^2\)
    • \(\sqrt{50} \approx 7.071...\)
    • \(B^2 = 50\) (exactly)
    • Answer: 50
  3. With A = 5, B = 3, find \(A^2 - B^2\)
    • \(25 - 9 = 16\)
    • Answer: 16

Problem 17: Calculator: Formula Evaluation with CALC (x)

Use the CALC button to evaluate expressions with different variable values:

  1. Enter the compound-growth formula \(A \cdot (1 + B)^C\) (start value A, rate B as a decimal, C years). Calculate for A = 1200, B = 0.03, C = 4. Then for A = 1200, B = 0.03, C = 8. Then for A = 2600, B = 0.07, C = 5

  2. Enter the percentage-change formula \(\frac{y - x}{x} \cdot 100\) (old value x, new value y). Calculate for x = 80, y = 92. Then for x = 250, y = 215. Then for x = 1.6, y = 2

  3. Evaluate \(2x^2 + 3x - 5\) for x = 1, 2, 3, 4

  1. Compound growth \(A \cdot (1 + B)^C\)
    • Enter formula, press CALC, input values
    • A = 1200, B = 0.03, C = 4: 1350.61
    • A = 1200, B = 0.03, C = 8: 1520.12
    • A = 2600, B = 0.07, C = 5: 3646.63
    • Use CALC to quickly change A, B and C
  2. Percentage change \(\frac{y - x}{x} \cdot 100\)
    • x = 80, y = 92: 15, an increase of 15 %
    • x = 250, y = 215: \(-14\), a decrease of 14 %
    • x = 1.6, y = 2: 25, an increase of 25 %
    • Use CALC to quickly change x and y
  3. \(2x^2 + 3x - 5\)
    • x = 1: \(2 + 3 - 5 = 0\)
    • x = 2: \(8 + 6 - 5 = 9\)
    • x = 3: \(18 + 9 - 5 = 22\)
    • x = 4: \(32 + 12 - 5 = 39\)

Problem 18: Calculator: Chain Calculations with Ans (x)

Use the Ans button to perform chain calculations:

  1. Calculate compound interest: Start with 1000 EUR at 5% annual interest. Use Ans to find the value after 1, 2, 3, and 4 years.

  2. Heron’s method for \(\sqrt{5}\): Start with guess 2, then iterate using \(\frac{1}{2}(x + \frac{5}{x})\)

  1. Compound interest at 5%
    • Year 0: 1000
    • Year 1: 1000 × 1.05 = → 1050
    • Year 2: × 1.05 = (Ans used automatically) → 1102.50
    • Year 3: × 1.05 = → 1157.63
    • Year 4: × 1.05 = → 1215.51
  2. Heron’s method for \(\sqrt{5}\)
    • Start: 2 = (initial guess)
    • Iterate: (Ans + 5 ÷ Ans) ÷ 2 =
    • Step 1: 2.25
    • Step 2: 2.236111…
    • Step 3: 2.2360679…
    • Converges to \(\sqrt{5} \approx 2.2360679...\)

Problem 19: Calculator: Logarithm Evaluations (x)

Use your calculator to find:

  1. \(\log_{10}(100)\)

  2. \(\ln(e^3)\)

  3. \(\log_2(64)\)

  4. \(\log_5(125) + \log_3(27)\)

  1. \(\log_{10}(100)\)
    • \(10^2 = 100\), so \(\log_{10}(100) = 2\)
    • Answer: 2
  2. \(\ln(e^3)\)
    • By definition of natural log, \(\ln(e^3) = 3\)
    • Answer: 3
  3. \(\log_2(64)\)
    • \(2^6 = 64\), so \(\log_2(64) = 6\)
    • Answer: 6
  4. \(\log_5(125) + \log_3(27)\)
    • \(5^3 = 125\), so \(\log_5(125) = 3\)
    • \(3^3 = 27\), so \(\log_3(27) = 3\)
    • Answer: 6