Tasks 02-03 - Fractional, Radical, Exponential & Logarithmic Equations

Section 02: Equations & Problem-Solving Strategies

A: Skills

Problem 1: Domain Restrictions (x)

Give the domain restrictions of each equation before you solve it. Then solve the equation.

  1. \(\frac{5}{x-3} = 2\)

  2. \(\frac{x}{x+2} + \frac{3}{x-1} = 1\)

  3. \(\frac{2x}{x^2-4} = \frac{1}{x-2}\)

  4. \(\frac{1}{x} + \frac{1}{x^2} = \frac{1}{2}\)

  5. \(\frac{x+1}{x-3} = \frac{x-2}{x+3}\)

  1. \(\frac{5}{x-3} = 2\)
    • Domain restriction: \(x \neq 3\)
    • Solve: \(5 = 2(x-3)\)
    • \(5 = 2x - 6\)
    • \(11 = 2x\)
    • \(x = 5.5\) (valid, since \(5.5 \neq 3\))
  2. \(\frac{x}{x+2} + \frac{3}{x-1} = 1\)
    • Domain restrictions: \(x \neq -2, x \neq 1\)
    • LCD: \((x+2)(x-1)\)
    • \(x(x-1) + 3(x+2) = (x+2)(x-1)\)
    • \(x^2 - x + 3x + 6 = x^2 + x - 2\)
    • \(2x + 6 = x - 2\)
    • \(x = -8\) (valid)
  3. \(\frac{2x}{x^2-4} = \frac{1}{x-2}\)
    • Factor the denominator: \(x^2 - 4 = (x-2)(x+2)\)
    • Domain restrictions: \(x \neq 2, x \neq -2\)
    • Multiply both sides by \((x-2)(x+2)\): \(2x = x + 2\)
    • \(x = 2\), but \(x = 2\) is excluded by the domain restrictions
    • No solution
  4. \(\frac{1}{x} + \frac{1}{x^2} = \frac{1}{2}\)
    • Domain restriction: \(x \neq 0\)
    • LCD: \(2x^2\)
    • \(2x + 2 = x^2\)
    • \(x^2 - 2x - 2 = 0\)
    • \(x = \frac{2 \pm \sqrt{4+8}}{2} = 1 \pm \sqrt{3}\)
    • Both solutions valid (neither equals 0)
  5. \(\frac{x+1}{x-3} = \frac{x-2}{x+3}\)
    • Domain restrictions: \(x \neq 3, x \neq -3\)
    • Cross multiply: \((x+1)(x+3) = (x-2)(x-3)\)
    • \(x^2 + 4x + 3 = x^2 - 5x + 6\)
    • \(9x = 3\)
    • \(x = \frac{1}{3}\) (valid)

Problem 2: Solving Rational Equations (x)

Solve each equation. Give the domain restrictions first.

  1. \(\frac{4}{x} - \frac{3}{2x} = \frac{1}{4}\)

  2. \(\frac{x+2}{x-1} - \frac{x-1}{x+2} = 0\)

  3. \(\frac{2}{x-3} + \frac{x}{x+3} = \frac{12}{x^2-9}\)

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

  1. \(\frac{4}{x} - \frac{3}{2x} = \frac{1}{4}\)
    • Domain: \(x \neq 0\)
    • LCD: \(4x\)
    • \(16 - 6 = x\)
    • \(x = 10\) (valid)
  2. \(\frac{x+2}{x-1} - \frac{x-1}{x+2} = 0\)
    • Domain: \(x \neq 1, x \neq -2\)
    • \(\frac{x+2}{x-1} = \frac{x-1}{x+2}\)
    • Cross multiply: \((x+2)^2 = (x-1)^2\)
    • \(x^2 + 4x + 4 = x^2 - 2x + 1\)
    • \(6x = -3\)
    • \(x = -\frac{1}{2}\) (valid)
  3. \(\frac{2}{x-3} + \frac{x}{x+3} = \frac{12}{x^2-9}\)
    • Note: \(x^2 - 9 = (x-3)(x+3)\)
    • Domain: \(x \neq 3, x \neq -3\)
    • LCD: \((x-3)(x+3)\)
    • \(2(x+3) + x(x-3) = 12\)
    • \(2x + 6 + x^2 - 3x = 12\)
    • \(x^2 - x - 6 = 0\)
    • \((x-3)(x+2) = 0\)
    • \(x = 3\) (excluded by the domain) or \(x = -2\) (valid)
    • Solution: \(x = -2\)
  4. \(\frac{1}{x-2} + \frac{2}{x+1} = \frac{3}{x-2}\)
    • Domain: \(x \neq 2, x \neq -1\)
    • Subtract \(\frac{1}{x-2}\): \(\frac{2}{x+1} = \frac{3-1}{x-2} = \frac{2}{x-2}\)
    • Therefore: \(x + 1 = x - 2\)
    • \(1 = -2\) (impossible)
    • No solution

Problem 3: Radical Equations (xx)

Solve each equation. Squaring can create extraneous solutions (candidates that do not solve the original equation), so verify every candidate in the original equation.

  1. \(\sqrt{x + 5} = 3\)

  2. \(\sqrt{2x - 1} = x - 2\)

  3. \(\sqrt{x^2 + 3} = 2\)

  4. \(x = \sqrt{x + 6}\)

  5. \(\sqrt{4x + 1} = 2x - 1\)

  6. \(\sqrt{x + 8} = 2 + \sqrt{x}\)

  7. \(\sqrt{2x + 5} - \sqrt{x + 2} = 1\)

Hint: in f) and g), isolate one root first. In g) you have to square twice.

  1. \(\sqrt{x + 5} = 3\)
    • Square: \(x + 5 = 9\)
    • \(x = 4\)
    • Check: \(\sqrt{4 + 5} = \sqrt{9} = 3\) ✓
  2. \(\sqrt{2x - 1} = x - 2\)
    • Square: \(2x - 1 = x^2 - 4x + 4\)
    • \(x^2 - 6x + 5 = 0\)
    • \((x - 5)(x - 1) = 0\)
    • \(x = 5\) or \(x = 1\)
    • Check \(x = 5\): \(\sqrt{9} = 3\) and \(5 - 2 = 3\) ✓
    • Check \(x = 1\): \(\sqrt{1} = 1\) but \(1 - 2 = -1\) ✗
    • Solution: \(x = 5\)
  3. \(\sqrt{x^2 + 3} = 2\)
    • Square: \(x^2 + 3 = 4\)
    • \(x^2 = 1\)
    • \(x = \pm 1\)
    • Check both: \(\sqrt{1 + 3} = 2\) ✓
    • Solutions: \(x = 1\) or \(x = -1\)
  4. \(x = \sqrt{x + 6}\)
    • Square: \(x^2 = x + 6\)
    • \(x^2 - x - 6 = 0\)
    • \((x - 3)(x + 2) = 0\)
    • \(x = 3\) or \(x = -2\)
    • Check \(x = 3\): \(3 = \sqrt{9} = 3\) ✓
    • Check \(x = -2\): \(-2 \neq \sqrt{4} = 2\) ✗
    • Solution: \(x = 3\)
  5. \(\sqrt{4x + 1} = 2x - 1\)
    • Square: \(4x + 1 = 4x^2 - 4x + 1\)
    • \(4x^2 - 8x = 0\)
    • \(4x(x - 2) = 0\)
    • \(x = 0\) or \(x = 2\)
    • Check \(x = 0\): \(\sqrt{1} = 1\) but \(-1 \neq 1\) ✗
    • Check \(x = 2\): \(\sqrt{9} = 3\) and \(4 - 1 = 3\) ✓
    • Solution: \(x = 2\)
  6. \(\sqrt{x + 8} = 2 + \sqrt{x}\)
    • The root on the left is already isolated. Square: \(x + 8 = 4 + 4\sqrt{x} + x\)
    • \(4 = 4\sqrt{x}\)
    • \(\sqrt{x} = 1\)
    • \(x = 1\)
    • Check: \(\sqrt{9} = 3\) and \(2 + 1 = 3\) ✓
  7. \(\sqrt{2x + 5} - \sqrt{x + 2} = 1\)
    • Isolate: \(\sqrt{2x + 5} = 1 + \sqrt{x + 2}\)
    • Square: \(2x + 5 = 1 + 2\sqrt{x + 2} + x + 2\)
    • \(x + 2 = 2\sqrt{x + 2}\)
    • Square again: \((x + 2)^2 = 4(x + 2)\)
    • \((x + 2)(x + 2 - 4) = 0\)
    • \((x + 2)(x - 2) = 0\)
    • \(x = -2\) or \(x = 2\)
    • Check \(x = 2\): \(\sqrt{9} - \sqrt{4} = 3 - 2 = 1\) ✓
    • Check \(x = -2\): \(\sqrt{1} - \sqrt{0} = 1 - 0 = 1\) ✓
    • Solutions: \(x = -2\) or \(x = 2\)
    • Note: dividing \((x + 2)^2 = 4(x + 2)\) by \(x + 2\) loses the solution \(x = -2\).

Problem 4: Exponential Equations and Inequalities (x)

Solve each equation or inequality:

  1. \(2^{x+3} = 128\)

  2. \(5^x \cdot 25^{x-1} = 125\)

  3. \(3^{2x} - 12 \cdot 3^x + 27 = 0\)

  4. \(4^x - 2^{x+1} - 8 = 0\)

  5. \(2^{x+1} \cdot 5^x = 2000\)

  6. \(3^{x-2} \leq 9^{x+1}\)

Hint: For a positive base \(a\), the power \(a^x\) is positive for every \(x\).

  1. \(2^{x+3} = 128\)
    • Recognize: \(128 = 2^7\)
    • So: \(2^{x+3} = 2^7\)
    • Therefore: \(x + 3 = 7\)
    • Solution: \(x = 4\)
  2. \(5^x \cdot 25^{x-1} = 125\)
    • Rewrite: \(5^x \cdot (5^2)^{x-1} = 5^3\)
    • Simplify: \(5^x \cdot 5^{2x-2} = 5^3\)
    • Combine: \(5^{x + 2x - 2} = 5^3\)
    • So: \(3x - 2 = 3\)
    • Solution: \(x = \frac{5}{3}\)
  3. \(3^{2x} - 12 \cdot 3^x + 27 = 0\)
    • Let \(u = 3^x\): \(u^2 - 12u + 27 = 0\)
    • Factor: \((u - 3)(u - 9) = 0\)
    • So: \(3^x = 3\) or \(3^x = 9\)
    • Solutions: \(x = 1\) or \(x = 2\)
  4. \(4^x - 2^{x+1} - 8 = 0\)
    • Note: \(4^x = (2^2)^x = (2^x)^2\) and \(2^{x+1} = 2 \cdot 2^x\)
    • Let \(u = 2^x\): \(u^2 - 2u - 8 = 0\)
    • Factor: \((u - 4)(u + 2) = 0\)
    • Since \(u = 2^x > 0\), only \(u = 4\) is valid
    • So: \(2^x = 4 = 2^2\)
    • Solution: \(x = 2\)
  5. \(2^{x+1} \cdot 5^x = 2000\)
    • Split off the constant factor: \(2 \cdot 2^x \cdot 5^x = 2000\)
    • Combine the powers with the same exponent: \(2 \cdot (2 \cdot 5)^x = 2 \cdot 10^x = 2000\)
    • So: \(10^x = 1000 = 10^3\)
    • Solution: \(x = 3\)
  6. \(3^{x-2} \leq 9^{x+1}\)
    • Rewrite the right side: \(9^{x+1} = (3^2)^{x+1} = 3^{2x+2}\)
    • Inequality: \(3^{x-2} \leq 3^{2x+2}\)
    • The base 3 is greater than 1, so the exponents keep the direction: \(x - 2 \leq 2x + 2\)
    • \(-4 \leq x\)
    • Solution: \(x \geq -4\), i.e. \([-4; \infty)\)

Problem 5: Logarithmic Equations (xx)

Solve each equation. Give the domain restrictions first.

  1. \(\log_3(x + 4) = 2\)

  2. \(\log(x) + \log(x + 3) = 1\)

  3. \(\log_2(x) - \log_8(x) = 1\)

  4. \(\log_x(16) = 4\)

  5. \(3^{\log_3(x)} + x = 10\)

  6. \(\log_2(2^x + 1) = x - 1\)

Note: an equation can have several solutions or none.

Hint: A logarithm is only defined for a positive argument.

  1. \(\log_3(x + 4) = 2\)
    • Domain: \(x > -4\)
    • Convert: \(x + 4 = 3^2 = 9\)
    • Solution: \(x = 5\) (valid)
  2. \(\log(x) + \log(x + 3) = 1\)
    • Domain: \(x > 0\) (then also \(x + 3 > 0\))
    • Product rule: \(\log(x(x + 3)) = 1\)
    • So: \(x(x + 3) = 10\)
    • Expand: \(x^2 + 3x - 10 = 0\)
    • Factor: \((x + 5)(x - 2) = 0\)
    • Since \(x > 0\): \(x = 2\)
  3. \(\log_2(x) - \log_8(x) = 1\)
    • Domain: \(x > 0\)
    • Change of base: \(\log_8(x) = \frac{\log_2(x)}{\log_2(8)} = \frac{\log_2(x)}{3}\)
    • Substitute: \(\log_2(x) - \frac{\log_2(x)}{3} = 1\)
    • Simplify: \(\frac{2\log_2(x)}{3} = 1\)
    • So: \(\log_2(x) = \frac{3}{2}\)
    • Solution: \(x = 2^{3/2} = 2\sqrt{2}\)
  4. \(\log_x(16) = 4\)
    • Domain: \(x > 0, x \neq 1\) (a base must be positive and different from 1)
    • Convert: \(x^4 = 16\)
    • So: \(x^4 = 2^4\), i.e. \(x = 2\) or \(x = -2\)
    • Solution: \(x = 2\) (\(x = -2\) is excluded by the domain)
  5. \(3^{\log_3(x)} + x = 10\)
    • Domain: \(x > 0\)
    • Since \(3^{\log_3(x)} = x\) for \(x > 0\): \(x + x = 10\)
    • Solution: \(x = 5\)
    • Check: \(3^{\log_3(5)} + 5 = 5 + 5 = 10\) ✓
  6. \(\log_2(2^x + 1) = x - 1\)
    • Domain: \(2^x + 1 > 0\) holds for every \(x\)
    • Convert: \(2^x + 1 = 2^{x-1}\)
    • Multiply by 2: \(2 \cdot 2^x + 2 = 2^x\)
    • This gives: \(2^x + 2 = 0\), i.e. \(2^x = -2\)
    • Since \(2^x > 0\), this is impossible
    • No solution

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. “A soap manufactory’s total weekly costs for \(x\) bars of soap are \(C = 0.005x^2 + 2x + 450\) EUR. The average cost per bar (total costs divided by the number of bars) is 5.75 EUR.”

    Check: Do 150 bars fulfil this condition?

  2. “Two packing machines of a spice trader pack one delivery. Machine A alone needs 5 hours, machine B alone needs 3 hours. Working together, they need \(t\) hours.”

    Check: Is \(t = 4\), the average of 5 and 3, correct?

  3. “An engineering office bought a 3D printer for 1,600 EUR. The resale value of the printer (the price it would still fetch if sold) halves every 2 years.”

    Hint: the value is multiplied by 0.5 once every 2 years, so after \(t\) years it has been halved \(t/2\) times.

    Check: Is the printer worth 400 EUR after 3 years?

  4. “A software company with 3,200 employees introduces an online course on a new data protection rule. Each week, 25 % of the employees who have not yet completed the course complete it. The company wants at least 95 % of its employees to have completed the course.”

    Check: Are 10 weeks enough?

  5. “The equation \(\log_2(x - 3) + \log_2(x + 4) = 3\) is only defined for …”

    Check: The zero form of this equation has the solutions \(x = 4\) and \(x = -5\). Which of them solves the original equation?

  1. \(\frac{0.005x^2 + 2x + 450}{x} = 5.75\) with \(x \neq 0\) (in the context \(x > 0\)). Multiply by \(x\):

    \(0.005x^2 + 2x + 450 = 5.75x \iff 0.005x^2 - 3.75x + 450 = 0 \iff x^2 - 750x + 90000 = 0\) (multiply by 200).

    Check: yes, \(C = 0.005 \cdot 150^2 + 2 \cdot 150 + 450 = 862.50\) EUR and \(\frac{862.50}{150} = 5.75\) EUR. Factored, \(x^2 - 750x + 90000 = (x - 150)(x - 600)\), so 600 bars also have an average cost of 5.75 EUR.

    Typical mistake: \(0.005x^2 + 2x + 450 = 5.75\). This sets the total costs equal to the average cost. The average cost is the total costs divided by \(x\).

  2. Machine A packs \(\frac{1}{5}\) of the delivery per hour, machine B \(\frac{1}{3}\). Working together, these amounts add up:

    \(\frac{1}{5} + \frac{1}{3} = \frac{1}{t}\) with \(t \neq 0\), so \(\frac{8}{15} = \frac{1}{t}\) and \(t = \frac{15}{8} = 1.875\) hours (1 h 52.5 min).

    Check: no. Together the machines must be faster than the faster machine alone, so \(t < 3\).

    Typical mistake: adding or averaging the times. The amounts of work per hour add up, the times do not.

  3. With \(t\) in years: \(V(t) = 1600 \cdot 0.5^{t/2}\) EUR.

    Check: no, \(V(3) = 1600 \cdot 0.5^{1.5} \approx 565.69\) EUR.

    Typical mistake: linear thinking: “the printer loses 800 EUR in 2 years, i.e. 400 EUR per year, so \(1600 - 3 \cdot 400 = 400\) EUR”. Halving means multiplying by 0.5 every 2 years. Per year the factor is \(0.5^{1/2} \approx 0.7071\), a loss of about 29.3 % of the current value.

  4. Employees who have not yet completed the course after \(n\) weeks: \(3200 \cdot 0.75^n\). At least 95 % completed means at most 5 % not completed:

    \(3200 \cdot 0.75^n \leq 0.05 \cdot 3200 \iff 0.75^n \leq 0.05\)

    Equality \(0.75^n = 0.05\) gives \(n = \frac{\ln(0.05)}{\ln(0.75)} \approx 10.41\). The share \(0.75^n\) falls every week, so from then on it stays at most 5 %. So the company needs 11 weeks.

    Check: no. After 10 weeks, \(0.75^{10} \approx 0.0563\), i.e. about 180 employees (5.6 %) have not yet completed the course.

    Typical mistake: \(0.75^n \geq 0.95\). The term \(0.75^n\) is the share of employees who have not yet completed the course, so it has to fall to 5 %.

  5. Both logarithms need a positive argument: \(x - 3 > 0\) and \(x + 4 > 0\). Together: \(x > 3\), i.e. the equation is only defined for \(x \in (3; \infty)\).

    Check: only \(x = 4\). It lies in the domain, and \(\log_2(1) + \log_2(8) = 0 + 3 = 3\) ✓. \(x = -5\) lies outside the domain.

    Typical mistake: reading the domain off the combined equation \(\log_2\big((x - 3)(x + 4)\big) = 3\). At \(x = -5\) the product \((-8) \cdot (-1) = 8\) is positive, so the combined logarithm exists, but \(\log_2(-8)\) and \(\log_2(-1)\) in the original equation do not. The domain always comes from the original equation.

Problem 7: One Paragraph, All Conditions (xx)

A company organises a boat trip for its employees. The rent for the boat, 1,440 EUR, is shared equally by all participants. In addition, each participant pays 18 EUR for the buffet. The boat takes at most 60 people, and the trip only takes place if at least 24 employees sign up. The works council (the elected representatives of the employees) wants the price per participant to be at most 45 EUR. The organiser also wants to know for which number of participants the price per participant is exactly 42 EUR.

List every condition in this paragraph as a mathematical statement, with \(x\) the number of participants. Give the domain restriction of your price term and write the 42-EUR condition in zero form. Do not solve anything.

Check: Which of your conditions rules out 50 participants?

  1. Participants: \(x\) is a whole number
  2. Boat rent per participant: \(\frac{1440}{x}\) EUR, defined only for \(x \neq 0\)
  3. Price per participant: \(p = \frac{1440}{x} + 18\) EUR
  4. Capacity of the boat: \(x \leq 60\)
  5. Minimum number of participants: \(x \geq 24\)
  6. Works council: \(\frac{1440}{x} + 18 \leq 45\)
  7. Exactly 42 EUR: \(\frac{1440}{x} + 18 = 42\). Multiplying by \(x\) is allowed, since \(x \neq 0\): \(1440 + 18x = 42x \iff 24x - 1440 = 0\)

Together (conditions 1, 4 and 5): \(x \in \{24, 25, \ldots, 60\}\). All these values satisfy the domain restriction \(x \neq 0\).

Check: for \(x = 50\) the price is \(\frac{1440}{50} + 18 = 28.80 + 18 = 46.80\) EUR \(> 45\) EUR. Condition 6 (works council) rules out 50 participants.

Typical mistake: \(p = \frac{1440 + 18}{x}\). Every participant pays the 18 EUR for the buffet in full. Only the boat rent is shared.

C: Applied

Problem 8: Stocktaking at a DIY Store (xx)

Once a year, a DIY store (a shop for tools and building materials) counts all the goods in stock (stocktaking). Its own staff team alone would need \(x\) hours for this. An external service team is 16 hours faster than the own team. Working together, the two teams need 15 hours.

  1. Explain why the own team counts \(\frac{1}{x}\) of the stock per hour. Set up an equation for the joint work and give its domain restrictions and the condition that the situation adds.

(for verification: \(\frac{1}{x} + \frac{1}{x - 16} = \frac{1}{15}\))

  1. Solve the equation. Decide which solution fits the situation and give the times the two teams would need alone.

(for verification: 40 hours and 24 hours)

  1. One hour of the own team costs 180 EUR, one hour of the service team 330 EUR. Compute the costs of the stocktaking if only the own team counts, if only the service team counts and if both teams count together. Interpret the result.

  2. The store can close for at most 12 hours. A second service team joins the two teams. Determine how many hours the second service team may need at most on its own so that all three teams together finish within 12 hours.

Hint: the teams finish within 12 hours if together they count at least \(\frac{1}{12}\) of the stock per hour. The denominator is positive, so you can multiply both sides by it without changing the sign.

a) In \(x\) hours the own team counts the whole stock once. At a constant speed it counts \(\frac{1}{x}\) of the stock per hour. The service team needs \(x - 16\) hours and counts \(\frac{1}{x - 16}\) of the stock per hour. Together they count \(\frac{1}{15}\) of the stock per hour:

\[\frac{1}{x} + \frac{1}{x - 16} = \frac{1}{15}\]

Domain restrictions: \(x \neq 0\) and \(x \neq 16\). The situation adds: both times are positive, so \(x - 16 > 0\), i.e. \(x > 16\).

b) Multiply by the common denominator \(15x(x - 16)\):

\[15(x - 16) + 15x = x(x - 16) \iff 30x - 240 = x^2 - 16x \iff x^2 - 46x + 240 = 0\]

\(\Delta = 46^2 - 4 \cdot 240 = 2116 - 960 = 1156 = 34^2\):

\[x = \frac{46 \pm 34}{2}, \qquad x_1 = 40, \quad x_2 = 6\]

Both values satisfy the domain restrictions, so both solve the equation. But \(x = 6\) violates \(x > 16\): the service team would need \(6 - 16 = -10\) hours. So \(x = 40\).

The own team needs 40 hours alone, the service team 24 hours.

Check: \(\frac{1}{40} + \frac{1}{24} = \frac{3}{120} + \frac{5}{120} = \frac{8}{120} = \frac{1}{15}\) ✓

c)

  • Own team alone: \(40 \cdot 180 = 7{,}200\) EUR
  • Service team alone: \(24 \cdot 330 = 7{,}920\) EUR
  • Both together: \(15 \cdot (180 + 330) = 15 \cdot 510 = 7{,}650\) EUR

The own team alone is the cheapest option, but the store would be closed for 40 hours. Both teams together finish after 15 hours for 450 EUR more. The service team alone is slower than both teams together and also more expensive, so it is never the best choice.

d) Let \(y\) be the time of the second service team alone in hours, \(y > 0\). Together the three teams have to count at least \(\frac{1}{12}\) of the stock per hour:

\[\frac{1}{40} + \frac{1}{24} + \frac{1}{y} \geq \frac{1}{12} \iff \frac{1}{15} + \frac{1}{y} \geq \frac{1}{12} \iff \frac{1}{y} \geq \frac{1}{12} - \frac{1}{15} = \frac{5 - 4}{60} = \frac{1}{60}\]

Multiply by \(60y > 0\) (the sign stays): \(60 \geq y\).

The second service team may need at most 60 hours on its own.

Check with \(y = 60\): \(\frac{3}{120} + \frac{5}{120} + \frac{2}{120} = \frac{10}{120} = \frac{1}{12}\) ✓

Problem 9: Fixed-Term Deposits of a Car Dealership (xx)

A car dealership put its surplus cash (money it does not need for its business at the moment) into two fixed-term deposits (money that stays in the account for a fixed time at a fixed interest rate) for one year: \(x\) thousand EUR at bank A and 2,000 EUR more at bank B. After the year, bank A paid 240 EUR interest and bank B paid 420 EUR. The interest rate of bank B was 1 percentage point higher than that of bank A (e.g. 5 % instead of 4 %).

  1. Give the interest rates of both banks as fractions with \(x\), and the domain restrictions. Show that \(x\) satisfies \(x^2 - 16x + 48 = 0\).

Hint: interest rate = interest ÷ amount. Write all amounts in thousand EUR, e.g. 240 EUR = 0.24 thousand EUR.

  1. Solve the equation and give the amounts and interest rates of both banks for each solution. The dealership had at most 15,000 EUR of surplus cash. Decide which solution describes the deposits.

(for verification: 4,000 EUR at 6 % and 6,000 EUR at 7 %)

  1. Compute the interest rate for the whole amount of 10,000 EUR. Explain why it is not 6.5 %.

  2. After the year, the dealership puts the whole balance of 10,660 EUR into a new deposit with yearly compounding. It wants to have 12,000 EUR after two years. Determine the interest rate it needs, in percent rounded to two decimal places.

a) Bank A: \(\frac{0.24}{x}\). Bank B holds \(x + 2\) thousand EUR: \(\frac{0.42}{x + 2}\).

Domain restrictions: \(x \neq 0\) and \(x \neq -2\). In the context also \(x > 0\).

One percentage point is \(0.01\):

\[\frac{0.42}{x + 2} - \frac{0.24}{x} = 0.01\]

Multiply by \(100x(x + 2)\):

\[42x - 24(x + 2) = x(x + 2) \iff 18x - 48 = x^2 + 2x \iff x^2 - 16x + 48 = 0 \quad \checkmark\]

b) Factors of 48 with sum \(-16\): \(-4\) and \(-12\).

\[(x - 4)(x - 12) = 0 \iff x = 4 \text{ or } x = 12\]

Both values satisfy the domain restrictions:

  • \(x = 4\): bank A 4,000 EUR at \(\frac{0.24}{4} = 6\,\%\), bank B 6,000 EUR at \(\frac{0.42}{6} = 7\,\%\)
  • \(x = 12\): bank A 12,000 EUR at \(\frac{0.24}{12} = 2\,\%\), bank B 14,000 EUR at \(\frac{0.42}{14} = 3\,\%\)

Both solutions fit the text so far (rates 1 percentage point apart). But \(x = 12\) needs \(12{,}000 + 14{,}000 = 26{,}000\) EUR, more than the 15,000 EUR available. So the deposits were 4,000 EUR at 6 % and 6,000 EUR at 7 %.

c) Total interest: \(240 + 420 = 660\) EUR on \(10{,}000\) EUR:

\[\frac{660}{10000} = 0.066 = 6.6\,\%\]

It is not the average \(6.5\,\%\) of the two rates, because the higher rate applies to the larger amount: 60 % of the money earns 7 %, only 40 % earns 6 %. Indeed \(0.4 \cdot 6\,\% + 0.6 \cdot 7\,\% = 6.6\,\%\).

d) With \(q = 1 + r\) the growth factor per year:

\[10660 \cdot q^2 = 12000 \iff q^2 = \frac{12000}{10660} \approx 1.125704 \iff q = \pm\sqrt{1.125704} \approx \pm 1.060992\]

A negative growth factor would turn the positive balance negative, so \(q \approx 1.060992\) and \(r \approx 6.10\,\%\).

Check: \(10660 \cdot 1.061^2 \approx 12{,}000.19\) EUR ✓

Problem 10: Sales of Printed Road Atlases (xx)

A map publisher sells printed road atlases. Since navigation apps have become common, its yearly sales fall continuously. The publisher models them by \(S(t) = 40000 \cdot e^{-kt}\) atlases per year, \(t\) years after the start of the observation, with a constant \(k > 0\). As in the formula \(A = Pe^{rt}\) for continuous compounding, a negative rate in the exponent, here \(-k\), describes a decrease. Two years after the start, the publisher sold 32,400 atlases per year.

  1. Determine \(k\) exactly and rounded to four decimal places. Show that \(e^{-k} = 0.9\) and interpret this value for the yearly sales.

(for verification: \(k = \ln\left(\frac{10}{9}\right) \approx 0.1054\))

  1. Determine the time in which the yearly sales halve, in years rounded to one decimal place.

  2. A new edition only pays off while the yearly sales are at least 8,000 atlases. Determine until when this is the case, in years rounded to one decimal place.

  3. At the start, the publisher sells an atlas for 15 EUR, and it raises the price by 4 % every year. Show that the yearly revenue from the atlases is \(R(t) = 600000 \cdot 0.936^t\) EUR. Determine after how many years the revenue has halved and compare the result with b).

a) \(S(2) = 32400\):

\[40000 \cdot e^{-2k} = 32400 \iff e^{-2k} = 0.81 \iff -2k = \ln(0.81) \iff k = -\frac{\ln(0.81)}{2}\]

Since \(0.81 = 0.9^2\), \(\ln(0.81) = 2\ln(0.9)\), so \(k = -\ln(0.9) = \ln\left(\frac{10}{9}\right) \approx 0.1054\).

Then \(e^{-k} = e^{\ln(0.9)} = 0.9\) ✓

Interpretation: \(S(t + 1) = 40000 \cdot e^{-k(t+1)} = S(t) \cdot e^{-k} = 0.9 \cdot S(t)\). Every year the sales fall by 10 % of the previous year’s sales.

b) Half of the sales at any time:

\[e^{-kt} = 0.5 \iff -kt = \ln(0.5) \iff t = \frac{\ln(2)}{k} = \frac{\ln(2)}{\ln(10/9)} \approx 6.58\]

The yearly sales halve every 6.6 years.

c)

\[40000 \cdot e^{-kt} \geq 8000 \iff e^{-kt} \geq 0.2 = e^{\ln(0.2)}\]

The base \(e\) is greater than 1, so the exponents keep the direction:

\[-kt \geq \ln(0.2) = -\ln(5) \iff t \leq \frac{\ln(5)}{k} \approx 15.28\]

(Dividing by \(-k < 0\) flips the sign.) A new edition pays off for about 15.3 years after the start.

Check with \(S(t) = 40000 \cdot 0.9^t\) from a): \(S(15) \approx 8{,}236\) atlases, \(S(16) \approx 7{,}412\) atlases ✓

d) With a) the sales are \(S(t) = 40000 \cdot (e^{-k})^t = 40000 \cdot 0.9^t\). The price after \(t\) years is \(15 \cdot 1.04^t\) EUR. Revenue = quantity × price:

\[R(t) = 40000 \cdot 0.9^t \cdot 15 \cdot 1.04^t = 600000 \cdot (0.9 \cdot 1.04)^t = 600000 \cdot 0.936^t \quad \checkmark\]

Halved revenue:

\[0.936^t = 0.5 \iff t = \frac{\ln(0.5)}{\ln(0.936)} \approx 10.48\]

The revenue halves after about 10.5 years, about 3.9 years later than the sales. The yearly price increases make up for part of the falling sales.

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: New Washing Machines for a Laundrette [22 pts. total]

A laundrette (a self-service laundry) near a university replaces its old washing machines. A customer pays 5.95 EUR per wash. This gross price includes 19 % VAT (value added tax) on the net price: gross price = net price + 19 % of the net price. The laundrette keeps only the net price. With a standard machine, water, energy and detergent cost the laundrette 2.00 EUR per wash (running costs).

(Earnings per wash = net price − running costs. A machine has paid for itself once its earnings add up to its price. With the same earnings every month, the payback period in months is the price of the machine divided by its earnings per month.)

Part a)

Compute the net price and the earnings per wash of a standard machine. A standard machine costs 9,000 EUR, and the owner wants a new machine to pay for itself within 18 months. Determine how many washes per month a standard machine needs at least for this. [4 pts.]

(for verification: earnings of 3.00 EUR per wash)

Net price. The gross price is 119 % of the net price:

\[\frac{5.95}{1.19} = 5.00 \text{ EUR}\]

(Subtracting 19 % of the gross price, \(5.95 \cdot 0.81 \approx 4.82\) EUR, uses the wrong base.)

Earnings per wash: \(5.00 - 2.00 = 3.00\) EUR.

Washes per month. With \(n\) washes per month, a standard machine earns \(3n\) EUR per month. Within 18 months its earnings have to cover the price:

\[18 \cdot 3n \geq 9000 \iff 54n \geq 9000 \iff n \geq 166.67\]

Only whole washes are possible, so round up: at least 167 washes per month. With 166 washes the earnings are only \(18 \cdot 3 \cdot 166 = 8{,}964\) EUR.

Grading: 1 Pt. net price with the net price as the base, 1 Pt. earnings per wash, 1 Pt. inequality, 1 Pt. rounding up to 167 washes with reason.

Part b)

The laundrette can also buy an eco model for 9,720 EUR. It needs less water and energy: with the same number of washes, it earns 90 EUR more per month than a standard machine, and it pays for itself 2 months earlier. Let \(x\) be the monthly earnings of a standard machine in EUR.

Set up an equation for the payback periods, give its domain restrictions and show that \(x\) satisfies

\[x^2 + 450x - 405000 = 0\]

[5 pts.]

Payback periods in months: standard machine \(\frac{9000}{x}\), eco model \(\frac{9720}{x + 90}\). The eco model needs 2 months less:

\[\frac{9000}{x} - \frac{9720}{x + 90} = 2\]

Domain restrictions: the denominators must not be zero, so \(x \neq 0\) and \(x \neq -90\). In the context also \(x > 0\).

Multiply by the common denominator \(x(x + 90)\):

\[9000(x + 90) - 9720x = 2x(x + 90)\]

\[9000x + 810000 - 9720x = 2x^2 + 180x\]

\[0 = 2x^2 + 900x - 810000 \quad \big| : 2\]

\[x^2 + 450x - 405000 = 0 \quad \checkmark\]

Grading: 1 Pt. both payback periods, 1 Pt. equation, 1 Pt. domain restrictions, 1 Pt. multiplying by the common denominator, 1 Pt. simplifying to the zero form.

Part c)

Solve the equation and decide with a reason which of its solutions fits the situation. Give the payback periods of both models and the number of washes per month. Decide which model meets the owner’s target from part a). [5 pts.]

(for verification: \(x = 450\))

Quadratic formula. \(\Delta = 450^2 - 4 \cdot 1 \cdot (-405000) = 202500 + 1620000 = 1822500 = 1350^2\):

\[x_{1,2} = \frac{-450 \pm 1350}{2}, \qquad x_1 = 450, \quad x_2 = -900\]

Which solution fits. Both values satisfy the domain restrictions, so both solve the equation. But \(x = -900\) would mean that a standard machine loses 900 EUR every month. Such a machine never pays for itself, and a payback period of \(\frac{9000}{-900} = -10\) months has no meaning. So \(x = 450\) EUR per month.

Payback periods: standard machine \(\frac{9000}{450} = 20\) months, eco model \(\frac{9720}{450 + 90} = \frac{9720}{540} = 18\) months. Check: \(20 - 18 = 2\) ✓

Washes per month: \(\frac{450}{3.00} = 150\) washes.

Decision: the standard machine misses the target of 18 months: it pays for itself only after 20 months, since 150 washes are fewer than the 167 washes from a). The eco model meets the target with exactly 18 months.

Grading: 1 Pt. solving the quadratic equation, 1 Pt. rejecting \(x = -900\) with reason, 1 Pt. both payback periods, 1 Pt. 150 washes, 1 Pt. decision with the target from a).

Part d)

Compute the running costs per wash of the eco model. Compute by how many percent they are lower than those of a standard machine, and by how many percent the earnings per wash of the eco model are higher. Explain why the two percentages differ. [3 pts.]

Eco model: it earns \(\frac{540}{150} = 3.60\) EUR per wash, so its running costs are \(5.00 - 3.60 = 1.40\) EUR per wash.

Running costs:

\[\frac{1.40 - 2.00}{2.00} = -0.30 = -30\,\%\]

Earnings per wash:

\[\frac{3.60 - 3.00}{3.00} = 0.20 = 20\,\%\]

Explanation: both changes are the same 0.60 EUR per wash, but the base differs. 0.60 EUR are 30 % of the old running costs of 2.00 EUR, but only 20 % of the old earnings of 3.00 EUR.

Grading: 1 Pt. running costs of 1.40 EUR, 1 Pt. both percentages, 1 Pt. explanation with the base.

Part e)

Half of the running costs of each model are energy costs. The other half, for water and detergent, stays the same, and so does the net price. Energy prices rise by 7 % per year.

Give a term for the running costs per wash of a standard machine \(t\) years from now. Determine when the earnings per wash of a standard machine have fallen to 2.00 EUR, in years rounded to one decimal place, and explain why this time is the doubling time of the energy costs. Compute the earnings per wash of the eco model at that time. [5 pts.]

Term. Today a standard machine uses 1.00 EUR of energy and 1.00 EUR of water and detergent per wash. Only the energy costs grow, by the factor 1.07 per year:

\[K(t) = 1.00 + 1.00 \cdot 1.07^t \text{ EUR}\]

Earnings of 2.00 EUR. The earnings per wash are \(5.00 - K(t) = 4.00 - 1.07^t\) EUR:

\[4.00 - 1.07^t = 2.00 \iff 1.07^t = 2 \iff t = \frac{\ln(2)}{\ln(1.07)} \approx 10.2\]

The earnings per wash fall to 2.00 EUR after about 10.2 years.

Explanation. Earnings of 2.00 EUR mean running costs of 3.00 EUR, i.e. energy costs of \(3.00 - 1.00 = 2.00\) EUR. That is twice today’s energy costs of 1.00 EUR, so the time is exactly the doubling time of the energy costs.

Eco model at that time. Its energy costs double from 0.70 EUR to 1.40 EUR, and water and detergent still cost 0.70 EUR. Running costs \(0.70 + 1.40 = 2.10\) EUR, earnings \(5.00 - 2.10 = 2.90\) EUR per wash. The advantage of the eco model grows from 0.60 EUR to 0.90 EUR per wash.

Grading: 1 Pt. term for the running costs, 1 Pt. equation for earnings of 2.00 EUR, 1 Pt. solving with logarithms, 1 Pt. explanation with the doubling of the energy costs, 1 Pt. earnings of the eco model.

Calculator Practice

Problem 11: Calculator: Cubic Equations (xx)

Solve each equation by hand: find one solution by trying small whole numbers or by grouping in pairs, then factor the rest. Use the CALC key to evaluate the left side at your trial values and to check every solution.

  1. \(2x^3 - 3x^2 - 11x + 6 = 0\)

  2. \(x^3 - 3x^2 - 4x + 12 = 0\)

Hint: If \(x = a\) is a solution, the left side is \((x - a)\) times a quadratic factor. Write this factor with unknown coefficients, multiply out and compare the coefficients with the left side.

Enter the left side, press CALC, enter a value for \(x\) and press =. A result of 0 means that the value is a solution.

  1. \(2x^3 - 3x^2 - 11x + 6 = 0\)
    • Trial values with CALC: \(x = 0\) gives 6, \(x = 1\) gives \(-6\), \(x = -1\) gives 12, \(x = 2\) gives \(-12\), \(x = -2\) gives 0 ✓
    • So \((x + 2)\) is a factor. The other factor has the form \(2x^2 + bx + 3\), since \(x \cdot 2x^2 = 2x^3\) and \(2 \cdot 3 = 6\)
    • Expand: \((x + 2)(2x^2 + bx + 3) = 2x^3 + (b + 4)x^2 + (2b + 3)x + 6\)
    • Compare the \(x^2\) terms: \(b + 4 = -3\), so \(b = -7\) (check the \(x\) terms: \(2 \cdot (-7) + 3 = -11\) ✓)
    • \((x + 2)(2x^2 - 7x + 3) = 0\)
    • Quadratic formula for \(2x^2 - 7x + 3 = 0\): \(\Delta = 49 - 24 = 25\), \(x = \frac{7 \pm 5}{4}\), so \(x = 3\) or \(x = \frac{1}{2}\)
    • Solutions: \(x = -2\), \(x = \frac{1}{2}\), \(x = 3\)
    • Check with CALC: \(x = 0.5\) and \(x = 3\) both give 0 ✓
  2. \(x^3 - 3x^2 - 4x + 12 = 0\)
    • Group in pairs: \(x^2(x - 3) - 4(x - 3) = 0\)
    • Factor out \((x - 3)\): \((x - 3)(x^2 - 4) = 0\)
    • Difference of squares: \((x - 3)(x - 2)(x + 2) = 0\)
    • Solutions: \(x = -2\), \(x = 2\), \(x = 3\)
    • Check with CALC: \(x = -2\), \(x = 2\) and \(x = 3\) all give 0 ✓

Problem 12: Calculator: SOLVE Function with Newton’s Method (xx)

Use the SOLVE function (SHIFT + CALC) to find solutions numerically:

  1. Find one solution of \(x^3 - 5x + 3 = 0\) (start with the initial guess \(x = 1\)).

  2. Solve \(2^x = x + 4\). Start once with the initial guess \(x = 3\) and once with \(x = -4\). Explain why you need the calculator here.

  3. Solve \(2x^2 - 14 = 0\) and give the exact solutions.

  1. \(x^3 - 5x + 3 = 0\)
    • Enter the equation, press SHIFT + CALC
    • Initial guess: 1
    • Answer: \(x \approx 0.6566\) (one of three solutions)
    • Start with \(x = -3\) to find another solution: \(x \approx -2.491\)
  2. \(2^x = x + 4\)
    • Enter the equation, press SHIFT + CALC, enter the initial guess and press =
    • Initial guess 3: \(x \approx 2.7562\)
    • Initial guess \(-4\): \(x \approx -3.9346\)
    • Check: \(2^{2.7562} \approx 6.756\) and \(2.7562 + 4 = 6.7562\) ✓
    • The unknown appears in an exponent and in a linear term at the same time. Neither a common base, nor logarithms, nor a substitution turns this equation into one we can solve by hand, so a numerical method is needed. The two initial guesses lead to two different solutions.
  3. \(2x^2 - 14 = 0\)
    • Enter the equation, initial guess 1 gives \(x \approx 2.6458\), initial guess \(-1\) gives \(x \approx -2.6458\)
    • Exact: \(2x^2 = 14 \Rightarrow x^2 = 7\)
    • Answers: \(x = \pm\sqrt{7} \approx \pm 2.6458\)