Tasks 04-03 - Exponential & Logarithmic Functions

Section 04: Advanced Functions

A: Skills

Problem 1: Basic Exponential Evaluation (x)

Consider the exponential function \(f(x) = 3 \cdot 2^x\).

  1. Calculate \(f(0)\), \(f(2)\) and \(f(-1)\) without a calculator.

  2. Find the \(y\)-intercept of the function.

  3. Decide whether the function describes growth or decay.

  4. State the domain and the range of \(f\) and describe what happens to \(f(x)\) for \(x \to -\infty\).

  5. Write \(f\) in the form \(f(x) = 3 \cdot e^{kx}\). Give \(k\) exactly and rounded to four decimal places.

  6. Compare \(f\) with the power function \(p(x) = x^4\): compute both functions at \(x = 5\), \(x = 10\) and \(x = 20\) and decide which one is larger for large \(x\).

  1. Insert the values:

    • \(f(0) = 3 \cdot 2^0 = 3 \cdot 1 = 3\)
    • \(f(2) = 3 \cdot 2^2 = 3 \cdot 4 = 12\)
    • \(f(-1) = 3 \cdot 2^{-1} = 3 \cdot \frac{1}{2} = 1.5\)

  2. The \(y\)-intercept is the value at \(x = 0\): \(f(0) = 3\), so the \(y\)-intercept is \((0, 3)\). Every function \(a \cdot b^x\) passes through \((0, a)\).

  3. The base \(b = 2\) is greater than 1, so \(f\) describes growth: the value doubles when \(x\) grows by 1.

  4. Domain: all real numbers. \(2^x > 0\) for every \(x\), so \(f(x) = 3 \cdot 2^x > 0\). Range: \((0; \infty)\).

    For \(x \to -\infty\), \(2^x = \frac{1}{2^{-x}}\) becomes a tiny positive number, e.g. \(f(-10) = \frac{3}{1024} \approx 0.0029\). So \(f(x) \to 0\): the graph approaches the horizontal asymptote \(y = 0\) (the \(x\)-axis), but never reaches it.

  5. Write the base as a power of \(e\): \(2 = e^{\ln(2)}\), so \(2^x = \left(e^{\ln(2)}\right)^x = e^{\ln(2) \cdot x}\).

    \[f(x) = 3 \cdot e^{\ln(2) \cdot x}, \qquad k = \ln(2) \approx 0.6931\]

    Check: \(f(1) = 3 \cdot e^{\ln(2)} = 3 \cdot 2 = 6\) ✓. In general, \(b = e^k\) and \(k = \ln(b)\).

  6. Values:

\(x\) 5 10 20
\(f(x) = 3 \cdot 2^x\) 96 3,072 3,145,728
\(p(x) = x^4\) 625 10,000 160,000

At \(x = 5\) and \(x = 10\), the power function is larger, but between \(x = 10\) and \(x = 20\) the exponential function overtakes it. For large \(x\), \(f\) is larger: when \(x\) grows by 1, \(f\) always doubles, while \(p\) grows by a smaller and smaller factor. An exponential function with \(b > 1\) eventually grows faster than every power function.

Problem 2: Identifying Exponential Functions (x)

Decide which of the following are exponential functions. For each exponential function, write it in the form \(a \cdot b^x\), give the initial value \(a\) and the base \(b\), and decide whether it describes growth or decay.

  1. \(g(x) = 5^{2x}\)

  2. \(h(x) = 4x^3\)

  3. \(k(x) = 4 \cdot (0.7)^x\)

  4. \(m(x) = 2^x + 3\)

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

  6. \(p(x) = 6 \cdot e^{-0.2x}\)

  7. \(q(x) = 5 \cdot 4^{-x/2}\)

An exponential function has the variable in the exponent and a fixed positive base \(b \neq 1\). Growth for \(b > 1\), decay for \(0 < b < 1\).

  1. \(g(x) = 5^{2x} = (5^2)^x = 25^x\): exponential, \(a = 1\), \(b = 25\), growth.

  2. \(h(x) = 4x^3\): not exponential. The variable is the base and the exponent is fixed, so this is a power function.

  3. \(k(x) = 4 \cdot (0.7)^x\): exponential, \(a = 4\), \(b = 0.7\), decay (\(0 < 0.7 < 1\)).

  4. \(m(x) = 2^x + 3\): not of the form \(a \cdot b^x\). It is the exponential function \(2^x\) shifted up by 3, so its graph approaches \(y = 3\) instead of \(y = 0\).

  5. \(n(x) = 3^{x + 2} = 3^2 \cdot 3^x = 9 \cdot 3^x\): exponential, \(a = 9\), \(b = 3\), growth. Check: \(n(0) = 3^2 = 9\) ✓.

  6. \(p(x) = 6 \cdot e^{-0.2x} = 6 \cdot \left(e^{-0.2}\right)^x\): exponential, \(a = 6\), \(b = e^{-0.2} \approx 0.8187\), decay.

  7. \(q(x) = 5 \cdot 4^{-x/2} = 5 \cdot \left(4^{-1/2}\right)^x = 5 \cdot \left(\frac{1}{2}\right)^x\): exponential, \(a = 5\), \(b = 0.5\), decay.

Problem 3: Logarithmic Properties (xx)

Simplify each expression with the logarithm laws. Assume \(x > 0\).

  1. \(\log_3(27x^2)\)

  2. \(\ln(e^{2x} \cdot \sqrt{x})\)

  3. \(2\log_5(5x) - \log_5(x^2)\)

  4. \(\log_2(8) + \log_2(x/4)\)

  1. \(\log_3(27x^2)\)
    • Product rule: \(\log_3(27x^2) = \log_3(27) + \log_3(x^2)\)
    • \(27 = 3^3\) and power rule: \(= 3 + 2\log_3(x)\)
  2. \(\ln(e^{2x} \cdot \sqrt{x})\)
    • Product rule: \(= \ln(e^{2x}) + \ln(\sqrt{x})\)
    • \(\ln(e^{u}) = u\) and \(\sqrt{x} = x^{1/2}\): \(= 2x + \frac{1}{2}\ln(x)\)
  3. \(2\log_5(5x) - \log_5(x^2)\)
    • First term: \(2\log_5(5x) = 2\left(\log_5(5) + \log_5(x)\right) = 2 + 2\log_5(x)\)
    • Second term: \(\log_5(x^2) = 2\log_5(x)\)
    • Together: \(2 + 2\log_5(x) - 2\log_5(x) = 2\)
  4. \(\log_2(8) + \log_2(x/4)\)
    • \(\log_2(8) = \log_2(2^3) = 3\)
    • Quotient rule: \(\log_2(x/4) = \log_2(x) - \log_2(4) = \log_2(x) - 2\)
    • Together: \(3 + \log_2(x) - 2 = 1 + \log_2(x)\)

Problem 4: Logarithmic and Exponential Equations (xx)

Solve the following equations. Give each solution exactly and rounded to three decimal places.

  1. \(\log_3(x+8) + \log_3(x) = 2\)

  2. \(5 \cdot e^{0.4x} - 3 = 17\)

  3. \(e^{2x} - 7e^x + 10 = 0\)

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

Hint for c): \(e^{2x} = (e^x)^2\). Substitute \(u = e^x\).

  1. Domain: \(x + 8 > 0\) and \(x > 0\), so \(x > 0\).

    • Product rule: \(\log_3\big(x(x + 8)\big) = 2\)
    • Exponential form: \(x(x + 8) = 3^2 = 9 \iff x^2 + 8x - 9 = 0\)
    • Factor: \((x + 9)(x - 1) = 0\), so \(x = -9\) or \(x = 1\)
    • \(x = -9\) is not in the domain. Solution: \(x = 1\)

    Check: \(\log_3(9) + \log_3(1) = 2 + 0 = 2\) ✓

  2. Isolate the power first: \[5e^{0.4x} = 20 \iff e^{0.4x} = 4 \iff 0.4x = \ln(4) \iff x = \frac{\ln(4)}{0.4} = 5\ln(2) \approx 3.466\]

    Check: \(5 \cdot e^{0.4 \cdot 5\ln(2)} - 3 = 5 \cdot e^{2\ln(2)} - 3 = 5 \cdot 4 - 3 = 17\) ✓

  3. With \(u = e^x\): \[u^2 - 7u + 10 = 0 \iff (u - 2)(u - 5) = 0 \iff u = 2 \text{ or } u = 5\]

    Both values are positive, so both are possible values of \(e^x\):

    • \(e^x = 2 \iff x = \ln(2) \approx 0.693\)
    • \(e^x = 5 \iff x = \ln(5) \approx 1.609\)

  4. Take the logarithm of both sides and use the power rule: \[x\ln(3) = (x + 1)\ln(2) \iff x\ln(3) - x\ln(2) = \ln(2) \iff x\big(\ln(3) - \ln(2)\big) = \ln(2)\] \[x = \frac{\ln(2)}{\ln(3) - \ln(2)} = \frac{\ln(2)}{\ln(1.5)} \approx 1.710\]

    Check: \(3^{1.710} \approx 6.54\) and \(2^{2.710} \approx 6.54\) ✓

Problem 5: Logarithmic Functions as Inverses (xx)

  1. Determine the inverse function of \(f(x) = 2 \cdot e^{0.5x}\). State the domain and the range of \(f\) and of its inverse function. Check your inverse function with the value \(f(2)\).

  2. Consider \(g(x) = \log_3(x + 9) - 3\). Describe how the graph of \(g\) arises from the graph of \(y = \log_3(x)\). State the domain and the range of \(g\) and determine its vertical asymptote.

  3. Determine the zero and the \(y\)-intercept of \(g\). The graph of \(y = \log_3(x)\) passes through \((1, 0)\). Give the corresponding point on the graph of \(g\).

  1. Solve \(y = 2e^{0.5x}\) for \(x\): \[\frac{y}{2} = e^{0.5x} \iff 0.5x = \ln\left(\frac{y}{2}\right) \iff x = 2\ln\left(\frac{y}{2}\right)\]

    Swap the names of the variables: \(f^{-1}(x) = 2\ln\left(\frac{x}{2}\right)\).

    • \(f\): domain all real numbers, range \((0; \infty)\)
    • \(f^{-1}\): domain \((0; \infty)\), range all real numbers (domain and range swap)

    Check: \(f(2) = 2e^{1} = 2e \approx 5.437\), and \(f^{-1}(2e) = 2\ln(e) = 2\) ✓

  2. \(x + 9\) instead of \(x\): the graph moves 9 units to the left. \(-3\) at the end: it moves 3 units down.

    • Domain: the logarithm needs \(x + 9 > 0\), so \(x > -9\). Domain: \((-9; \infty)\)
    • Range: all real numbers (moving the graph down does not change this)
    • Vertical asymptote \(x = -9\): for \(x \to -9\) from the right, \(x + 9\) becomes a tiny positive number. \(3^y\) is that small only for very negative \(y\), so \(\log_3(x + 9) \to -\infty\) and \(g(x) \to -\infty\). This is the mirror image of the horizontal asymptote \(y = 0\) of \(3^x\), moved 9 units to the left.

  3. Zero: \(\log_3(x + 9) = 3 \iff x + 9 = 3^3 = 27 \iff x = 18\). The zero is \((18, 0)\).

    • \(y\)-intercept: \(g(0) = \log_3(9) - 3 = 2 - 3 = -1\), so \((0, -1)\)
    • \((1, 0)\) moves 9 units to the left and 3 units down: \((-8, -3)\). Check: \(g(-8) = \log_3(1) - 3 = -3\) ✓

B: Set Up & Check

Problem 6: From Text to Functions (xx)

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

  1. “The number of passengers at a regional airport grows by 2.8 % per year.” A consultant describes the same airport with “a continuous growth rate of 2.8 % per year”. Give both models, with \(P_0\) passengers today and \(t\) in years.

    Check: The rule of 70 estimates a doubling time of \(\frac{70}{2.8} = 25\) years for both models. Compute the growth factor of both models after 25 years. Have the passengers doubled?

  2. “The number of landline connections of a telecom provider halves every 5 years.”

    Check: Does the number of landline connections fall by 10 % per year?

  3. “The number of companies that use a start-up’s accounting software doubles every 8 months.”

    Check: Does the number of companies triple within one year?

  4. “A savings bank pays 2.4 % interest per year on a company’s savings account, compounded monthly (each month, one twelfth of the yearly rate is added).”

    Check: Is the balance after one year exactly 2.4 % higher than today?

  5. “A city’s new parking app has \(N(t) = \frac{6000}{1 + 11e^{-0.3t}}\) users \(t\) months after its launch. The number of users can never exceed 6,000.” Translate the last sentence into a statement about the graph of \(N\).

    Check: Compute \(N(0)\) and \(N(24)\). Is there a time \(t\) with \(N(t) = 6000\)?

  1. Growth by 2.8 % per year multiplies the number by 1.028 every year. A continuous growth rate uses the base \(e\):

    \[P_1(t) = P_0 \cdot 1.028^t, \qquad P_2(t) = P_0 \cdot e^{0.028t}\]

    The second model grows by the factor \(e^{0.028} \approx 1.0284\) per year, i.e. by about 2.84 %.

    Check: \(1.028^{25} \approx 1.9945\), so with yearly growth the passengers have not quite doubled. \(e^{0.028 \cdot 25} = e^{0.7} \approx 2.0138\), so with continuous growth they have. The rule of 70 is an estimate. The exact doubling times are \(\frac{\ln(2)}{\ln(1.028)} \approx 25.10\) and \(\frac{\ln(2)}{0.028} \approx 24.76\) years.

    Typical mistake: \(P_0 \cdot e^{1.028t}\). The growth factor 1.028 does not belong in the exponent of \(e\): this model would multiply the passengers by \(e^{1.028} \approx 2.8\) every year.

  2. With \(C_0\) landline connections today and \(t\) in years: \(C(t) = C_0 \cdot 0.5^{t/5}\).

    Check: no. The yearly factor is \(0.5^{1/5} \approx 0.8706\), so the number falls by about 12.94 % per year.

    Typical mistake: “50 % in 5 years, so 10 % per year”. With 10 % per year, \(0.9^5 \approx 0.5905\) of the connections would be left after 5 years, i.e. 59 % and not half. Each yearly loss refers to what is left.

  3. With \(N_0\) companies today and \(t\) in months: \(N(t) = N_0 \cdot 2^{t/8}\).

    Check: no. \(N(12) = N_0 \cdot 2^{12/8} = N_0 \cdot 2^{1.5} = 2\sqrt{2} \cdot N_0 \approx 2.83 \cdot N_0\).

    Typical mistake: “12 months are 1.5 doubling periods, so the number grows by the factor \(1.5 \cdot 2 = 3\).” Doubling periods multiply the number: the factor is \(2^{1.5}\), not \(1.5 \cdot 2\).

  4. The monthly rate is \(\frac{0.024}{12} = 0.002\), and interest is added 12 times per year. With \(K_0\) EUR today and \(t\) in years:

    \[K(t) = K_0 \cdot \left(1 + \frac{0.024}{12}\right)^{12t} = K_0 \cdot 1.002^{12t}\]

    Check: no, slightly more. \(1.002^{12} \approx 1.024266\), i.e. about 2.43 %: from the second month on, the interest also earns interest.

    Typical mistake: \(K_0 \cdot 1.024^{12t}\). This adds the full yearly rate every month: after one year \(1.024^{12} \approx 1.329\), i.e. almost 33 % interest.

  5. \(N(t) < 6000\) for every \(t \geq 0\), and \(N(t) \to 6000\) for \(t \to \infty\): the graph of \(N\) has the horizontal asymptote \(y = 6000\). In the logistic model, 6,000 is the carrying capacity \(L\).

    Check: \(N(0) = \frac{6000}{1 + 11} = 500\) users and \(N(24) = \frac{6000}{1 + 11e^{-7.2}} \approx 5{,}951\) users. \(N(t) = 6000\) would need \(1 + 11e^{-0.3t} = 1\), i.e. \(11e^{-0.3t} = 0\). This is impossible because \(e^{-0.3t} > 0\). The denominator is always greater than 1, so \(N(t)\) stays below 6,000.

    Typical mistake: reading “can never exceed 6,000” as “reaches 6,000 at some time and then stays there”. The number of users gets closer and closer to 6,000, but never reaches it.

Problem 7: One Paragraph, All Conditions (xx)

A youth hostel puts its reserves of 30,000 EUR into a savings account at 1.8 % interest per year, compounded monthly. It wants to renovate its kitchen for 34,000 EUR from this account within at most 6 years. This year, the hostel has 12,000 overnight stays, and the owners expect the number of overnight stays to grow with a continuous growth rate of 5 % per year. The hostel has 40 beds and is open every day of the year.

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

Check: An assistant writes the balance as \(K(t) = 30000 \cdot 1.018^{12t}\) and the overnight stays as \(S(t) = 12000 \cdot 1.05^t\). Name the two mistakes.

  1. Balance in EUR: \(K(t) = 30000 \cdot \left(1 + \frac{0.018}{12}\right)^{12t} = 30000 \cdot 1.0015^{12t}\)
  2. Renovation: \(K(t) \geq 34000\)
  3. Deadline: \(0 \leq t \leq 6\)
  4. Overnight stays per year: \(S(t) = 12000 \cdot e^{0.05t}\)
  5. Capacity: at most \(40 \cdot 365 = 14{,}600\) overnight stays per year, so \(S(t) \leq 14600\) (the model can only hold as long as this is true)

Check: 1.018 is the growth factor for interest added once a year. With monthly compounding, the monthly factor is \(1 + \frac{0.018}{12} = 1.0015\). And a continuous growth rate of 5 % gives \(e^{0.05t}\), not \(1.05^t\).

Typical mistake: forgetting the capacity. The 40 beds limit the overnight stays to 14,600 per year, however fast the demand grows.

C: Applied

Problem 8: Court Bookings at a New Padel Centre (xx)

A padel centre (padel is a racket sport similar to tennis, played in pairs on a small court) opened in a city. The table shows the number of court bookings per week, \(t\) weeks after the opening:

Weeks after the opening \(t\) 0 4 8 12
Bookings per week \(N\) 160 240 360 540

Round logarithms, growth factors and \(k\) to four decimal places, percentages to two decimal places and times to one decimal place.

  1. Compute the factor by which the bookings grow in each 4-week period and explain why this shows exponential growth. Determine the weekly growth factor and the weekly growth rate in percent.

(for verification: weekly growth factor \(\approx 1.1067\))

  1. Compute \(\ln(N)\) for the four values in the table. Show that \(\ln(N)\) grows by the same amount every 4 weeks, and explain why the points \((t, \ln(N))\) lie on a straight line in a semi-log plot (a plot with a logarithmic scale on the vertical axis). Determine the slope of this line per week.

  2. Write the model as \(N(t) = 160 \cdot e^{kt}\) and explain how \(k\) is related to the slope from b). Show that \(N(t + 4) = 1.5 \cdot N(t)\) for every \(t \geq 0\). Determine the doubling time (the time after which the bookings have doubled) exactly and rounded. Compare it with the estimate of the rule of 70, using the weekly growth rate from a).

Hint for c): the logarithm laws give \(\ln\left(160 \cdot e^{kt}\right) = \ln(160) + kt\).

  1. The model describes how many bookings the customers ask for. The centre has 8 courts, and each court can be booked 10 times a day, 7 days a week. Determine the capacity (the maximum possible number of bookings) per week and after how many weeks the customers ask for more bookings than the centre can offer.

Hint for d): The larger of two positive numbers has the larger logarithm, so taking the logarithm of both sides keeps the inequality sign.

  1. The owner uses the model for a whole year: “After 52 weeks, our customers will ask for more than 30,000 bookings per week.” Check this number and explain why the model cannot describe the bookings for that long. Name the model from the lecture that describes growth with a limit and describe its three phases for the padel centre.

a) \(\frac{240}{160} = 1.5\), \(\frac{360}{240} = 1.5\) and \(\frac{540}{360} = 1.5\). In every 4-week period, the bookings are multiplied by the same factor 1.5. A constant factor in equal time steps is exponential growth: \(N(t) = 160 \cdot 1.5^{t/4}\).

Weekly growth factor \(b\): four weeks multiply the bookings by \(b^4 = 1.5\), so

\[b = 1.5^{1/4} = \sqrt[4]{1.5} \approx 1.1067\]

The bookings grow by about 10.67 % per week (not by \(\frac{50\,\%}{4} = 12.5\,\%\)).

b)

\(t\) 0 4 8 12
\(\ln(N)\) 5.0752 5.4806 5.8861 6.2916

The differences are all \(0.4055\). By the quotient rule, \(\ln(240) - \ln(160) = \ln\left(\frac{240}{160}\right) = \ln(1.5) \approx 0.4055\), and the same holds for the other steps. In a semi-log plot, the vertical axis shows \(\ln(N)\). It grows by the same amount in equal time steps, just like a linear function, so the points lie on a straight line with the slope

\[\frac{\ln(1.5)}{4} \approx \frac{0.4055}{4} \approx 0.1014 \text{ per week}\]

c) The models \(160 \cdot e^{kt}\) and \(160 \cdot 1.5^{t/4}\) agree if \(e^k = 1.5^{1/4}\), i.e.

\[k = \ln\left(1.5^{1/4}\right) = \frac{\ln(1.5)}{4} \approx 0.1014, \qquad N(t) = 160 \cdot e^{0.1014t}\]

With the hint, \(\ln(N) = \ln(160) + kt\) is a straight line in \(t\) with the slope \(k\) and the intercept \(\ln(160)\). So \(k\) is the slope from b), the continuous growth rate per week.

Every 4 weeks: replace \(t\) by \(t + 4\) and split the power with \(e^{u + v} = e^u \cdot e^v\):

\[N(t + 4) = 160 \cdot e^{k(t + 4)} = 160 \cdot e^{kt} \cdot e^{4k} = e^{4k} \cdot N(t)\]

From a), \(e^{4k} = \left(e^k\right)^4 = \left(1.5^{1/4}\right)^4 = 1.5\), so \(N(t + 4) = 1.5 \cdot N(t)\) for every \(t \geq 0\) ✓. The factor 1.5 from the table holds for every 4-week period, not only for the periods that start at \(t = 0\), 4 and 8.

Doubling time: \(160 \cdot e^{kt} = 320 \iff e^{kt} = 2 \iff kt = \ln(2)\):

\[t = \frac{\ln(2)}{k} = \frac{4\ln(2)}{\ln(1.5)} \approx 6.8 \text{ weeks}\]

Rule of 70 with 10.67 % per week: \(\frac{70}{10.67} \approx 6.6\) weeks. The estimate is a bit too low: the rule of 70 works well for small growth rates, and 10.67 % per week is a large rate.

d) Capacity: \(8 \cdot 10 \cdot 7 = 560\) bookings per week. Condition:

\[160 \cdot 1.5^{t/4} > 560 \iff 1.5^{t/4} > 3.5 \iff \frac{t}{4} \cdot \ln(1.5) > \ln(3.5) \iff t > \frac{4\ln(3.5)}{\ln(1.5)} \approx 12.4\]

The logarithm keeps the inequality sign (the larger of two positive numbers has the larger logarithm), and \(\ln(1.5) > 0\). After about 12.4 weeks, the customers ask for more bookings than the centre can offer. Check: \(N(12) = 540 < 560\) and \(N(13) = 160 \cdot 1.5^{3.25} \approx 598 > 560\) ✓.

e) \(N(52) = 160 \cdot 1.5^{52/4} = 160 \cdot 1.5^{13} \approx 31{,}139\) bookings per week. With the model, the owner’s number is right.

But the model cannot hold for that long: the centre can offer only 560 bookings per week, and a city has only a limited number of people who play padel. An exponential function grows without bound, and 10.67 % more every week cannot go on for a year.

Logistic growth \(P(t) = \frac{L}{1 + Ae^{-kt}}\) has a carrying capacity \(L\), the largest number of bookings the city’s players will ask for. Its three phases:

  1. Slow start: only a few people know the new centre.
  2. Rapid growth: word of mouth brings many new players, as in the first 12 weeks.
  3. Saturation: most padel players in the city already book regularly, and the bookings approach \(L\).

Problem 9: Saving for Electric Karts (xx)

A go-kart track (an indoor race track where customers drive small racing cars called karts) wants to replace its petrol karts with electric karts. Today, it puts 80,000 EUR into a savings account. Three banks make an offer:

  • Bank A: 3.2 % interest per year, added once a year
  • Bank B: 3.16 % interest per year, compounded monthly (each month, one twelfth of the yearly rate is added)
  • Bank C: 3.14 % interest per year, compounded continuously

Round growth factors to six decimal places and amounts of money to two decimal places. Keep the unrounded calculator values for further calculations.

  1. Give the balance after \(t\) years for each bank and compute the three balances after 6 years.

(for verification: Bank A: 96,642.50 EUR)

  1. Compute the growth factor per year of each offer. Explain why the order of the three banks is the same for every savings period, and decide which bank the go-kart track should choose.

  2. The adviser of Bank A says: “In 6 years, you earn \(6 \cdot 3.2\,\% = 19.2\,\%\) interest.” Explain the adviser’s mistake and compute the correct percentage.

  3. The electric karts cost 92,000 EUR today, and their price rises by 1 % per year. Determine after how many whole years the balance at the bank from b) is enough to buy the karts.

a) Balances in EUR after \(t\) years:

  • Bank A: \(K_A(t) = 80000 \cdot 1.032^t\)
  • Bank B: \(K_B(t) = 80000 \cdot \left(1 + \frac{0.0316}{12}\right)^{12t}\)
  • Bank C: \(K_C(t) = 80000 \cdot e^{0.0314t}\)

After 6 years:

  • \(K_A(6) = 80000 \cdot 1.032^6 \approx 96{,}642.50\) EUR
  • \(K_B(6) = 80000 \cdot \left(1 + \frac{0.0316}{12}\right)^{72} \approx 96{,}677.18\) EUR
  • \(K_C(6) = 80000 \cdot e^{0.1884} \approx 96{,}585.31\) EUR

b) Growth factors per year (for Bank B with the power rule \(a^{12t} = \left(a^{12}\right)^t\)):

  • Bank A: \(q_A = 1.032\)
  • Bank B: \(q_B = \left(1 + \frac{0.0316}{12}\right)^{12} \approx 1.032062\)
  • Bank C: \(q_C = e^{0.0314} \approx 1.031898\)

Every balance has the form \(80000 \cdot q^t\). For every \(t > 0\), the larger growth factor gives the larger balance: \(\frac{K_B(t)}{K_A(t)} = \left(\frac{q_B}{q_A}\right)^t > 1\) because \(\frac{q_B}{q_A} > 1\). So the order B, A, C holds for every savings period.

The go-kart track should choose Bank B, although its rate of 3.16 % is lower than Bank A’s 3.2 %: with monthly compounding, the interest of each month earns interest in the following months. Bank C compounds continuously, but its rate is too low to make up for it.

c) 19.2 % would be simple interest \(A = P(1 + rt)\), where only the 80,000 EUR earn interest. With compound interest, the interest of each year earns interest in the following years:

\[1.032^6 \approx 1.208031, \quad \text{i.e. } +20.80\,\%\]

That is \(96{,}642.50 - 80{,}000 = 16{,}642.50\) EUR of interest instead of \(80{,}000 \cdot 0.192 = 15{,}360\) EUR.

d) Price after \(t\) years: \(92000 \cdot 1.01^t\) EUR. Condition with Bank B:

\[80000 \cdot q_B^t \geq 92000 \cdot 1.01^t \iff \frac{q_B^t}{1.01^t} \geq \frac{92000}{80000} \iff \left(\frac{q_B}{1.01}\right)^t \geq 1.15\]

Take the logarithm of both sides (the sign stays) and use the power rule. \(\ln\left(\frac{q_B}{1.01}\right) > 0\), so the sign stays when dividing:

\[t \geq \frac{\ln(1.15)}{\ln\left(\frac{q_B}{1.01}\right)} \approx \frac{0.1398}{0.0216} \approx 6.47\]

Only whole years count, so round up: after 7 years. Check:

  • \(t = 6\): balance \(\approx 96{,}677.18\) EUR, price \(92000 \cdot 1.01^6 \approx 97{,}659.85\) EUR (not enough)
  • \(t = 7\): balance \(\approx 99{,}776.82\) EUR, price \(\approx 98{,}636.45\) EUR ✓

Problem 10: Training New Staff at a Mail-Order Pharmacy (xx)

A mail-order pharmacy (a pharmacy that sends medicines and health products by post) trains new staff for its warehouse. The time a new employee needs for order picking (collecting all items of one order from the shelves) is modelled by

\[T(t) = 4 + a \cdot e^{-kt}\]

in minutes per order, where \(t\) is the number of working days since the employee started and \(a, k > 0\). Experienced employees need 4 minutes per order. On the first day (\(t = 0\)), a new employee needs 12 minutes per order, after 5 working days only 7 minutes. Round \(k\) to four decimal places and times to two decimal places.

  1. Determine \(a\) and \(k\). Give \(k\) exactly and rounded.

(for verification: \(T(t) = 4 + 8e^{-kt}\) with \(k = \frac{1}{5}\ln\left(\frac{8}{3}\right) \approx 0.1962\))

  1. A team leader says: “In the first 5 days, the time fell by 5 minutes. So after 10 days, a new employee needs only 2 minutes per order.” Compute \(T(10)\) without a calculator and explain the team leader’s mistake.

Hint: \(e^{-10k} = \left(e^{-5k}\right)^2\).

  1. A new employee counts as fully trained once the time per order is at most 4.5 minutes. Determine after how many whole working days this is the case, and check your answer with the day before.

  2. Investigate the behavior of \(T(t)\) for \(t \to \infty\) and give the horizontal asymptote. Explain why, in this model, a new employee never becomes quite as fast as the experienced staff.

  3. A shift has 480 working minutes. Compute how many complete orders a new employee picks per shift on day 0 and on day 15. Give the number of orders per shift that the model approaches in the long run.

Hint: dividing the same positive number by a larger number gives a smaller result.

a) First day: \(T(0) = 4 + a \cdot e^0 = 4 + a = 12\), so \(a = 8\).

After 5 days:

\[4 + 8e^{-5k} = 7 \iff 8e^{-5k} = 3 \iff e^{-5k} = \frac{3}{8}\]

Take the natural logarithm, \(\ln\left(\frac{3}{8}\right) = -\ln\left(\frac{8}{3}\right)\):

\[-5k = -\ln\left(\frac{8}{3}\right) \iff k = \frac{1}{5}\ln\left(\frac{8}{3}\right) \approx 0.1962\]

\[T(t) = 4 + 8e^{-0.1962t}\]

b) With the hint and \(e^{-5k} = \frac{3}{8}\) from a):

\[T(10) = 4 + 8 \cdot \left(\frac{3}{8}\right)^2 = 4 + 8 \cdot \frac{9}{64} = 4 + \frac{9}{8} = 5.125 \approx 5.13 \text{ minutes}\]

The team leader assumed a linear decrease of 1 minute per day. In the model, the extra time above the 4 minutes of experienced staff shrinks by the same factor \(\frac{3}{8}\) every 5 days: from 8 to 3 minutes, then from 3 to \(\frac{9}{8}\) minutes. The time can never fall below 4 minutes, so 2 minutes are impossible.

c) Condition:

\[4 + 8e^{-kt} \leq 4.5 \iff 8e^{-kt} \leq 0.5 \iff e^{-kt} \leq \frac{1}{16}\]

Take the logarithm (the sign stays): \(-kt \leq \ln\left(\frac{1}{16}\right) = -\ln(16)\). Dividing by \(-k < 0\) flips the sign:

\[t \geq \frac{\ln(16)}{k} = \frac{5\ln(16)}{\ln\left(\frac{8}{3}\right)} \approx 14.13\]

Round up to whole working days: after 15 working days. Check:

  • \(t = 14\): \(T(14) = 4 + 8e^{-14k} \approx 4.51 > 4.5\) (not yet)
  • \(t = 15\): \(T(15) = 4 + 8 \cdot \left(\frac{3}{8}\right)^3 = 4 + \frac{27}{64} \approx 4.42 \leq 4.5\) ✓

d) For \(t \to \infty\): \(-kt \to -\infty\), so \(e^{-kt} \to 0\) and

\[T(t) \to 4 + 8 \cdot 0 = 4\]

Horizontal asymptote: \(y = 4\). In the long run, a new employee needs almost exactly 4 minutes per order, like the experienced staff. But \(8e^{-kt} > 0\) for every \(t\), so \(T(t) > 4\): in the model the new employee stays a tiny bit slower. After 15 days the difference is already less than half a minute.

e) Orders per shift \(= \frac{480}{T(t)}\):

  • Day 0: \(\frac{480}{12} = 40\) orders
  • Day 15: \(\frac{480}{4.421875} \approx 108.55\), i.e. 108 complete orders
  • Long run: \(T(t)\) approaches 4 from above, so \(\frac{480}{T(t)}\) approaches \(\frac{480}{4} = 120\) orders per shift from below. Dividing 480 by a number slightly larger than 4 gives slightly less than 120.

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: Customer Hotline of an Electricity Supplier [21 pts. total]

In January 2026, a regional electricity supplier launched an online customer portal, where customers can, for example, report their meter readings themselves. Since then, fewer customers call its hotline. In January 2026 the hotline received 15,000 calls, in March 2026 only 10,800 calls. The number of calls per month is modelled by

\[N(t) = N_0 \cdot e^{-kt}, \qquad k > 0\]

where \(t\) is the time in months since January 2026 (\(t = 0\) stands for January 2026, \(t = 2\) for March 2026). Round \(k\) to four decimal places, and times and percentages to two decimal places.

Part a)

Determine \(N_0\) and \(k\). Give \(k\) exactly and rounded. [4 pts.]

(for verification: \(k = -\frac{1}{2}\ln(0.72) \approx 0.1643\))

Initial value: \(N(0) = N_0 \cdot e^0 = N_0 = 15000\).

Second data point (\(t = 2\)):

\[15000 \cdot e^{-2k} = 10800 \iff e^{-2k} = \frac{10800}{15000} = 0.72\]

Natural logarithm of both sides, \(\ln(e^u) = u\):

\[-2k = \ln(0.72) \iff k = -\frac{1}{2}\ln(0.72) = \frac{1}{2}\ln\left(\frac{25}{18}\right) \approx 0.1643\]

\(\ln(0.72) < 0\), so \(k > 0\) as required.

\[N(t) = 15000 \cdot e^{-0.1643t}\]

Grading: 1 Pt. \(N_0 = 15000\), 1 Pt. equation \(e^{-2k} = 0.72\), 1 Pt. logarithm, 1 Pt. \(k\) exactly and rounded.

Part b)

Determine the percentage by which the number of calls falls per month. Explain why this percentage is smaller than 16.43 %. [3 pts.]

Monthly factor: from one month to the next, \(N\) is multiplied by \(e^{-k}\). From a), \(\left(e^{-k}\right)^2 = e^{-2k} = 0.72\), so

\[e^{-k} = \sqrt{0.72} \approx 0.8485\]

The number of calls falls by \(1 - 0.8485 = 0.1515\), i.e. by about 15.15 % per month.

Explanation: \(k \approx 0.1643\) is the continuous decay rate. In the continuous model, the decrease acts all the time, and every small loss refers to the smaller number that is left. Over one month, the losses add up to less than 16.43 % of the number at the start of the month: \(1 - e^{-0.1643} \approx 0.1515 < 0.1643\).

Grading: 1 Pt. monthly factor \(e^{-k} \approx 0.8485\), 1 Pt. 15.15 % per month, 1 Pt. explanation.

Part c)

Determine the half-life of the number of calls (the time after which it has halved) exactly and rounded. Explain what the half-life means for the hotline. [3 pts.]

Condition: half of the starting value.

\[15000 \cdot e^{-kt} = 7500 \iff e^{-kt} = \frac{1}{2} \iff -kt = \ln\left(\frac{1}{2}\right) = -\ln(2)\]

\[t = \frac{\ln(2)}{k} = \frac{2\ln(2)}{\ln\left(\frac{25}{18}\right)} \approx 4.22 \text{ months}\]

Meaning: every 4.22 months, the number of calls halves, whatever month we start from: from 15,000 to 7,500 calls, then to 3,750 calls after another 4.22 months, and so on.

Grading: 1 Pt. equation \(e^{-kt} = \frac{1}{2}\), 1 Pt. exact and rounded half-life, 1 Pt. meaning.

Part d)

The supplier will close one of its two hotline sites in the first month in which the hotline receives fewer than 2,400 calls. Determine this month. [4 pts.]

Condition:

\[15000 \cdot e^{-kt} < 2400 \iff e^{-kt} < 0.16\]

Logarithm of both sides (the sign stays):

\[-kt < \ln(0.16)\]

Dividing by \(-k < 0\) flips the sign:

\[t > \frac{\ln(0.16)}{-k} = \frac{2\ln(6.25)}{\ln\left(\frac{25}{18}\right)} \approx 11.16\]

First whole month: \(t\) counts whole months, so the first month is \(t = 12\), i.e. January 2027.

Check with \(e^{-k} = \sqrt{0.72}\):

  • \(t = 11\) (December 2026): \(N(11) = 15000 \cdot 0.72^{5.5} \approx 2{,}463 > 2400\)
  • \(t = 12\) (January 2027): \(N(12) = 15000 \cdot 0.72^6 \approx 2{,}090 < 2400\) ✓

Grading: 1 Pt. inequality \(e^{-kt} < 0.16\), 1 Pt. logarithm with sign flip, 1 Pt. January 2027, 1 Pt. check with \(t = 11\) and \(t = 12\).

Part e)

Investigate the behavior of \(N(t)\) for \(t \to \infty\) and interpret the result. The head of customer service says: “In the long run, we can shut down the hotline completely.” Assess this statement. [3 pts.]

Behavior: for \(t \to \infty\), \(-kt \to -\infty\), so \(e^{-kt} \to 0\) and

\[N(t) = 15000 \cdot e^{-kt} \to 0\]

The graph has the horizontal asymptote \(y = 0\).

Interpretation: in the long run, the model predicts almost no calls. But \(e^{-kt} > 0\) for every \(t\), so the model never reaches 0 calls either.

Assessment: the statement goes too far. The model only says that the calls become very few, never that they stop. In practice, some customers will keep calling, e.g. customers without internet access or with problems the portal cannot solve. The number of calls will level off above 0, and the model is only reliable for the first months after the launch.

Grading: 1 Pt. \(e^{-kt} \to 0\) and \(N(t) \to 0\), 1 Pt. interpretation (never exactly 0), 1 Pt. assessment with a reason.

Part f)

A trainee describes the calls with a linear function \(L\) through the same two data points instead. Determine \(L(t)\) and the time at which \(L\) predicts no calls at all. In May 2026, the hotline received 7,810 calls. Decide which model describes the calls better, and explain why \(L\) cannot be used in the long run. [4 pts.]

Linear function through \((0, 15000)\) and \((2, 10800)\):

\[m = \frac{10800 - 15000}{2 - 0} = -2100, \qquad L(t) = 15000 - 2100t\]

The linear model loses 2,100 calls every month.

No calls: \(15000 - 2100t = 0 \iff t = \frac{15000}{2100} \approx 7.14\). The linear model predicts no calls at all after about 7.14 months. From September 2026 (\(t = 8\)) on, it even gives negative numbers.

May 2026 (\(t = 4\)):

  • exponential model: \(N(4) = 15000 \cdot 0.72^2 = 7{,}776\) calls (34 calls off)
  • linear model: \(L(4) = 15000 - 8400 = 6{,}600\) calls (1,210 calls off)

The exponential model describes the calls much better.

Long run: the linear model loses the same number of calls every month, so it becomes negative after \(t \approx 7.14\), e.g. \(L(8) = -1800\) calls, which is impossible. The exponential model loses the same percentage every month, so its losses become smaller and smaller, and the number stays positive.

Grading: 1 Pt. slope and \(L(t)\), 1 Pt. zero with meaning, 1 Pt. comparison at \(t = 4\) with decision, 1 Pt. explanation (negative values, constant amount vs constant percentage).

Calculator Practice

Problem 11: Calculator: Exponential Calculations (x)

Calculate with your calculator and round to four decimal places. Use SHIFT + ln for \(e^x\), SHIFT + log for \(10^x\) and the x^ key for other powers.

  1. \(e^3\)

  2. \(e^{-0.4}\)

  3. \(10^{2.5}\)

  4. \(1.025^{16}\)

  5. \(e^{\ln(7)}\) (first ln, then SHIFT + ln)

  6. \(\frac{\ln(2)}{\ln(1.035)}\) with the ln key, and \(\log_{1.035}(2)\) with the log key as in Session 01-05. Explain why both give the same value and what this value means for money that grows by 3.5 % per year.

  1. \(e^3 \approx 20.0855\)

  2. \(e^{-0.4} \approx 0.6703\)

  3. \(10^{2.5} \approx 316.2278\)

  4. \(1.025^{16} \approx 1.4845\)

  5. \(e^{\ln(7)} = 7\): \(e^x\) and \(\ln(x)\) undo each other (the logarithm is the inverse function of \(e^x\))

  6. Both give about \(20.1488\). By the change of base formula, \(\log_{1.035}(2) = \frac{\ln(2)}{\ln(1.035)}\). It is the solution of \(1.035^t = 2\): money that grows by 3.5 % per year doubles in about 20.15 years.