Session 04-03 - Tasks

Exponential Functions Deep Dive

Exponential Functions - Problem Set

Problem 1: Basic Exponential Evaluation (x)

Given the exponential function \(f(x) = 3 \cdot 2^x\):

  1. Calculate \(f(0)\), \(f(2)\), and \(f(-1)\)

  2. Find the y-intercept of the function

  3. Determine if this represents growth or decay

Part a) Evaluating the function:

For \(f(0)\): \[f(0) = 3 \cdot 2^0 = 3 \cdot 1 = 3\]

For \(f(2)\): \[f(2) = 3 \cdot 2^2 = 3 \cdot 4 = 12\]

For \(f(-1)\): \[f(-1) = 3 \cdot 2^{-1} = 3 \cdot \frac{1}{2} = \frac{3}{2} = 1.5\]

Part b) Y-intercept:

The y-intercept occurs when \(x = 0\): \[f(0) = 3 \cdot 2^0 = 3\]

Therefore, the y-intercept is \((0, 3)\).

Part c) Growth or decay:

Since the base \(b = 2 > 1\), this represents exponential growth.

The function increases as \(x\) increases.

Problem 2: Identifying Exponential Functions (x)

Determine which of the following are exponential functions. If yes, identify the base and initial value:

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

  2. \(h(x) = x^5\)

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

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

Part a) \(g(x) = 5^{2x}\):

This can be rewritten as: \[g(x) = 5^{2x} = (5^2)^x = 25^x\]

This IS an exponential function with:

  • Base: \(25\)
  • Initial value: \(1\) (when written as \(1 \cdot 25^x\))

Part b) \(h(x) = x^5\):

This is a power function, NOT an exponential function.

  • In exponentials, the variable is in the exponent
  • Here, the variable is the base

Part c) \(k(x) = 4 \cdot (0.7)^x\):

This IS an exponential function with:

  • Base: \(0.7\)
  • Initial value: \(4\)
  • Since \(0 < 0.7 < 1\), this represents exponential decay

Part d) \(m(x) = 2^x + 3\):

This is NOT a pure exponential function.

  • It’s an exponential function with a vertical shift
  • The “+3” makes it not fit the standard form \(a \cdot b^x\)

Problem 3: Bacteria Growth Model (xx)

A bacteria culture starts with 500 cells and doubles every 3 hours.

  1. Write an exponential model for the number of bacteria after \(t\) hours

  2. How many bacteria will there be after 9 hours?

  3. How long will it take to reach 32,000 bacteria?

Part a) Creating the model:

Since the bacteria doubles every 3 hours:

  • Initial population: \(N_0 = 500\)
  • Growth factor: doubles (×2) every 3 hours
  • Model: \(N(t) = 500 \cdot 2^{t/3}\)

where \(t\) is time in hours.

Part b) Population after 9 hours:

\[N(9) = 500 \cdot 2^{9/3} = 500 \cdot 2^3 = 500 \cdot 8 = 4,000\]

After 9 hours, there will be 4,000 bacteria.

Part c) Time to reach 32,000:

Set up the equation: \[32,000 = 500 \cdot 2^{t/3}\]

Divide both sides by 500: \[64 = 2^{t/3}\]

Since \(64 = 2^6\): \[2^6 = 2^{t/3}\]

Therefore: \[6 = \frac{t}{3}\]

\[t = 18 \text{ hours}\]

The population will reach 32,000 bacteria after 18 hours.

Check: \(N(18) = 500 \cdot 2^{18/3} = 500 \cdot 2^6 = 500 \cdot 64 = 32,000\) ✓

Problem 4: Compound Interest Comparison (xx)

You want to invest €8,000 for 6 years. Compare these options:

  • Bank A: 4.5% annual interest, compounded quarterly
  • Bank B: 4.4% annual interest, compounded monthly
  • Bank C: 4.3% annual interest, compounded continuously

Which option yields the highest return?

Bank A: 4.5% quarterly compounding

Using \(A = P(1 + \frac{r}{n})^{nt}\):

  • Principal: \(P = 8000\)
  • Annual rate: \(r = 0.045\)
  • Compounds per year: \(n = 4\)
  • Time: \(t = 6\) years

Step 1: Calculate the quarterly rate \[\frac{r}{n} = \frac{0.045}{4} = 0.01125\]

Step 2: Calculate the number of periods \[nt = 4 \times 6 = 24 \text{ quarters}\]

Step 3: Calculate the growth factor \[A = 8000(1.01125)^{24}\]

Using a calculator: \((1.01125)^{24} = 1.3066\)

Step 4: Final amount \[A = 8000 \times 1.3066 = €10,452.80\]

Bank B: 4.4% monthly compounding

Using the same formula:

  • Principal: \(P = 8000\)
  • Annual rate: \(r = 0.044\)
  • Compounds per year: \(n = 12\)
  • Time: \(t = 6\) years

Step 1: Calculate the monthly rate \[\frac{r}{n} = \frac{0.044}{12} = 0.003667\]

Step 2: Calculate the number of periods \[nt = 12 \times 6 = 72 \text{ months}\]

Step 3: Calculate the growth factor \[A = 8000(1.003667)^{72}\]

Using a calculator: \((1.003667)^{72} = 1.3023\)

Step 4: Final amount \[A = 8000 \times 1.3023 = €10,418.40\]

Bank C: 4.3% continuous compounding

Using \(A = Pe^{rt}\):

  • Principal: \(P = 8000\)
  • Annual rate: \(r = 0.043\)
  • Time: \(t = 6\) years

Step 1: Calculate the exponent \[rt = 0.043 \times 6 = 0.258\]

Step 2: Calculate \(e^{0.258}\) Using a calculator: \(e^{0.258} = 1.2943\)

Step 3: Final amount \[A = 8000 \times 1.2943 = €10,354.40\]

Comparison Summary: | Bank | Rate | Compounding | Final Amount | |——|——|————-|————–| | A | 4.5% | Quarterly | €10,452.80 ✓ | | B | 4.4% | Monthly | €10,418.40 | | C | 4.3% | Continuous | €10,354.40 |

Bank A offers the best return. The higher interest rate (4.5%) more than compensates for the less frequent compounding.

Problem 5: Radioactive Decay (xx)

A radioactive isotope has a half-life of 8 days. Starting with 120 grams:

  1. Write the exponential decay model

  2. How much remains after 24 days?

  3. When will only 15 grams remain?

Part a) Decay model:

For half-life problems, we use: \[A(t) = A_0 \cdot (0.5)^{t/h}\]

where \(h\) is the half-life.

With \(A_0 = 120\) grams and \(h = 8\) days: \[A(t) = 120 \cdot (0.5)^{t/8}\]

Part b) Amount after 24 days:

\[A(24) = 120 \cdot (0.5)^{24/8}\] \[A(24) = 120 \cdot (0.5)^3\] \[A(24) = 120 \cdot 0.125\] \[A(24) = 15 \text{ grams}\]

After 24 days, 15 grams remain.

Part c) Time to reach 15 grams:

From part b), we already found this happens at 24 days, but let’s verify:

\[15 = 120 \cdot (0.5)^{t/8}\]

Divide by 120: \[0.125 = (0.5)^{t/8}\]

Since \(0.125 = \frac{1}{8} = (0.5)^3\): \[(0.5)^3 = (0.5)^{t/8}\]

Therefore: \[3 = \frac{t}{8}\] \[t = 24 \text{ days}\]

Only 15 grams will remain after 24 days.

Problem 6: Market Penetration Model (xx)

A new smartphone app launches with 1,000 users. The user base doubles every 2 months.

  1. Write the exponential growth model

  2. Project the number of users after 6 months

  3. How many users will there be after 10 months?

Part a) Growth model:

Since the user base doubles every 2 months: \[U(t) = 1000 \cdot 2^{t/2}\]

where \(t\) is time in months.

Part b) Users after 6 months:

\[U(6) = 1000 \cdot 2^{6/2} = 1000 \cdot 2^3 = 1000 \cdot 8 = 8,000\]

After 6 months, there will be 8,000 users.

Part c) Users after 10 months:

\[U(10) = 1000 \cdot 2^{10/2} = 1000 \cdot 2^5 = 1000 \cdot 32 = 32,000\]

After 10 months, there will be 32,000 users.

Business Interpretation: The exponential growth shows:

  • Month 0: 1,000 users (launch)
  • Month 2: 2,000 users (first doubling)
  • Month 4: 4,000 users
  • Month 6: 8,000 users
  • Month 8: 16,000 users
  • Month 10: 32,000 users

This rapid growth pattern is typical for viral apps in their early stages.

Problem 7: Comparing Investment Strategies (xxx)

You have €10,000 to invest. Compare three investment scenarios:

  1. Calculate the value of each investment after 4 years:

    • Option A: 6% annual interest, compounded annually
    • Option B: 5.8% annual interest, compounded semi-annually
    • Option C: 5.7% annual interest, compounded quarterly

  2. Option A grows by exactly 50% after how many full years? (Check: when does it first exceed €15,000?)

  3. If you need exactly €16,000 after 8 years, which option(s) will achieve this goal?

Part a) Value after 4 years:

Option A (6% annually): \[A = 10000(1.06)^4\] Calculate step by step:

  • \((1.06)^2 = 1.1236\)
  • \((1.06)^4 = 1.2625\) \[A = 10000 \cdot 1.2625 = €12,625\]

Option B (5.8% semi-annually): \[A = 10000(1 + \frac{0.058}{2})^{2 \cdot 4} = 10000(1.029)^8\] Calculate step by step:

  • \((1.029)^2 = 1.0588\)
  • \((1.029)^4 = 1.1215\)
  • \((1.029)^8 = 1.2578\) \[A = 10000 \cdot 1.2578 = €12,578\]

Option C (5.7% quarterly): \[A = 10000(1 + \frac{0.057}{4})^{4 \cdot 4} = 10000(1.01425)^{16}\] Calculate step by step:

  • \((1.01425)^4 = 1.0581\)
  • \((1.01425)^8 = 1.1196\)
  • \((1.01425)^{16} = 1.2536\) \[A = 10000 \cdot 1.2536 = €12,536\]

Ranking after 4 years: 1. Option A: €12,625 (best) 2. Option B: €12,578 3. Option C: €12,536

Part b) When Option A grows by 50%:

We need \(10000(1.06)^t = 15000\), so \((1.06)^t = 1.5\)

Check year by year:

  • Year 6: \((1.06)^6 = 1.4185\) → €14,185 (not yet)
  • Year 7: \((1.06)^7 = 1.5036\) → €15,036 (exceeds!)

Option A first exceeds €15,000 (50% growth) after 7 years.

Part c) Which options reach €16,000 after 8 years:

Option A: \(10000(1.06)^8 = 10000 \cdot 1.5938 = €15,938\) ❌ (falls short by €62)

Option B: \(10000(1.029)^{16} = 10000 \cdot 1.5825 = €15,825\) ❌ (falls short by €175)

Option C: \(10000(1.01425)^{32} = 10000 \cdot 1.5714 = €15,714\) ❌ (falls short by €286)

None of the options reach €16,000 after 8 years. Option A comes closest, missing the target by only €62.

Problem 8: COVID-19 Spread Analysis (xxxx)

During early 2020, COVID-19 cases in a region grew exponentially before interventions. The data shows:

  • Day 0: 64 confirmed cases
  • Cases doubled every 3 days initially
  • After day 12, strong interventions reduced the growth to 20% every 3 days
  • After day 24, lockdown reduced growth to 5% every 3 days

  1. Write the exponential models for each phase

  2. Calculate the number of cases at days 12, 24, and 30

  3. How many cases would there have been at day 30 without any interventions?

  4. By what factor did the interventions reduce the case count at day 30?

Part a) Exponential models for each phase:

Phase 1 (Days 0-12): Doubling every 3 days

  • Model: \(C_1(t) = 64 \cdot 2^{t/3}\) for \(0 \leq t \leq 12\)
  • At day 12: \(C(12) = 64 \cdot 2^{12/3} = 64 \cdot 2^4 = 64 \cdot 16 = 1,024\) cases

Phase 2 (Days 12-24): 20% growth every 3 days

  • Starting value: 1,024 cases (from Phase 1)
  • Model: \(C_2(t) = 1,024 \cdot (1.2)^{(t-12)/3}\) for \(12 < t \leq 24\)
  • At day 24: \(C(24) = 1,024 \cdot (1.2)^4 = 1,024 \cdot 2.0736 = 2,123\) cases

Phase 3 (Days 24-30): 5% growth every 3 days

  • Starting value: 2,123 cases (from Phase 2)
  • Model: \(C_3(t) = 2,123 \cdot (1.05)^{(t-24)/3}\) for \(t > 24\)

Part b) Cases at key days:

Day 12 (end of Phase 1): \[C(12) = 64 \cdot 2^{12/3} = 64 \cdot 2^4\] \[= 64 \cdot 16 = 1,024 \text{ cases}\]

Day 24 (end of Phase 2): Starting from 1,024 cases, with 20% growth every 3 days:

  • Number of 3-day periods: \((24-12)/3 = 4\)
  • Growth calculation: \(C(24) = 1,024 \cdot (1.2)^4\)

Step-by-step calculation of \((1.2)^4\):

  • \((1.2)^2 = 1.44\)
  • \((1.2)^4 = 1.44 \times 1.44 = 2.0736\)

Therefore: \(C(24) = 1,024 \cdot 2.0736 = 2,123 \text{ cases}\)

Day 30 (end of observation): Starting from 2,123 cases, with 5% growth every 3 days:

  • Number of 3-day periods: \((30-24)/3 = 2\)
  • Growth calculation: \(C(30) = 2,123 \cdot (1.05)^2\)
  • \((1.05)^2 = 1.1025\)

Therefore: \(C(30) = 2,123 \cdot 1.1025 = 2,341 \text{ cases}\)

Part c) Cases without interventions (continued doubling):

If the initial doubling pattern continued for all 30 days: \[C_{no-intervention}(30) = 64 \cdot 2^{30/3} = 64 \cdot 2^{10}\]

Calculate \(2^{10} = 1,024\): \[C_{no-intervention}(30) = 64 \cdot 1,024 = 65,536 \text{ cases}\]

Part d) Reduction factor:

Compare actual vs. no-intervention scenarios:

  • With interventions: 2,341 cases
  • Without interventions: 65,536 cases

\[\text{Reduction factor} = \frac{65,536}{2,341} \approx 28\]

The interventions reduced the case count by a factor of 28.

Public Health Insight:

  • With interventions: 2,341 cases
  • Without interventions: 65,536 cases
  • Lives saved: Potentially thousands

This demonstrates why “flattening the curve” through early intervention was crucial. The exponential growth would have overwhelmed healthcare systems without these measures.