Tasks 01-06 - Synthesis & Linear Equations

Integrating Mathematical Foundations

Problem 1: Multi-Technique Integration

Solve the following problems using multiple techniques:

  1. Solve for x: \(\log_2(x^2 - 9) = \log_2(x - 3) + 3\) (very hard and too early, sorry!)

  2. Factor completely, then evaluate at \(x = \sqrt{2}\): \(x^4 - 5x^2 + 4\)

  3. Simplify the expression \(\frac{(x^3 - 8) \cdot \sqrt{x + 2}}{(x - 2) \cdot (x^2 + 2x + 4)}\) when \(x = 6\)

  4. If \(2^x = 3\) and \(3^y = 4\), find the value of \(6^{xy}\)

  5. Expand \((1 + \sqrt{2})^3\) and express in the form \(a + b\sqrt{2}\)

  1. Solve: \(\log_2(x^2 - 9) = \log_2(x - 3) + 3\)
    • Rewrite: \(\log_2(x^2 - 9) = \log_2(x - 3) + \log_2(8)\)
    • \(\log_2(x^2 - 9) = \log_2[8(x - 3)]\)
    • Therefore: \(x^2 - 9 = 8(x - 3)\)
    • \(x^2 - 9 = 8x - 24\)
    • \(x^2 - 8x + 15 = 0\)
    • \((x - 3)(x - 5) = 0\)
    • So \(x = 3\) or \(x = 5\)
    • Check \(x = 3\): \(\log_2(0)\) is undefined, so \(x = 3\) is extraneous
    • Check \(x = 5\): \(\log_2(16) = \log_2(2) + 3 = 4 = 1 + 3\) ✓
    • Therefore: \(x = 5\)
  2. Factor: \(x^4 - 5x^2 + 4\)
    • Let \(u = x^2\): \(u^2 - 5u + 4 = (u-1)(u-4)\)
    • \(= (x^2 - 1)(x^2 - 4) = (x+1)(x-1)(x+2)(x-2)\)
    • At \(x = \sqrt{2}\): \((\sqrt{2}+1)(\sqrt{2}-1)(\sqrt{2}+2)(\sqrt{2}-2)\)
    • \(= (2-1)(2-4) = 1 \times (-2) = -2\)
  3. At \(x = 6\):
    • Numerator: \((216 - 8) \cdot \sqrt{8} = 208 \cdot 2\sqrt{2} = 416\sqrt{2}\)
    • Denominator: \(4 \cdot (36 + 12 + 4) = 4 \cdot 52 = 208\)
    • Result: \(\frac{416\sqrt{2}}{208} = 2\sqrt{2}\)
  4. From \(2^x = 3\): \(x = \log_2(3)\) From \(3^y = 4\): \(y = \log_3(4)\)
    • \(xy = \log_2(3) \cdot \log_3(4) = \log_2(4) = 2\)
    • Therefore: \(6^{xy} = 6^2 = 36\)
  5. \((1 + \sqrt{2})^3\)
    • Using binomial: \(1 + 3\sqrt{2} + 3(\sqrt{2})^2 + (\sqrt{2})^3\)
    • \(= 1 + 3\sqrt{2} + 6 + 2\sqrt{2}\)
    • \(= 7 + 5\sqrt{2}\)

Problem 2: Exponential-Logarithmic Systems

Solve the following systems and equations:

  1. System: \(2^x \cdot 3^y = 72\) and \(x = 3\) (find y, then verify with logarithms)

  2. If \(\log_a(b) = 2\) and \(\log_b(c) = 3\), find \(\log_a(c)\) and \(\log_c(a)\)

  3. A bacteria culture doubles every 3 hours. Starting with 1000 bacteria, the number after t hours = \(1000 \cdot 2^{t/3}\). After how many hours will there be exactly \(10^6\) bacteria?

  1. Given \(x = 3\) and \(2^x \cdot 3^y = 72\):
    • \(2^3 \cdot 3^y = 72\)
    • \(8 \cdot 3^y = 72\)
    • \(3^y = 9\)
    • \(y = 2\)
    • Verify: \(72 = 8 \times 9 = 2^3 \times 3^2\) ✓
  2. Given \(\log_a(b) = 2\) and \(\log_b(c) = 3\):
    • \(\log_a(c) = \log_a(b) \cdot \log_b(c) = 2 \times 3 = 6\)
    • \(\log_c(a) = \frac{1}{\log_a(c)} = \frac{1}{6}\)
  3. Solve \(10^6 = 1000 \cdot 2^{t/3}\):
    • \(1000 = 2^{t/3}\)
    • \(t/3 = \log_2(1000) = \frac{\ln(1000)}{\ln(2)} = 9.97\)
    • \(t = 29.9\) hours

Problem 3: Complex Integration Problem

A pharmaceutical company is developing a new drug with the following characteristics:

Concentration: After t hours, concentration = \(100 \cdot 2^{-t/4} + 20\) mg/L

Production Cost: Cost to produce x units = \(1000 + 50x + 0.1x^2\) euros

Market Demand: At price p euros, demand = \(10000 \cdot 0.95^p\) units

  1. Find when the concentration drops to half its initial value.

  2. At what time does the concentration reach exactly 30 mg/L?

  3. What is the production cost for 100 units? Express in scientific notation.

  4. If the company wants demand of exactly 5000 units, what price should they set?

  1. Initial concentration (t = 0): \(100 + 20 = 120\) mg/L Half value = 60 mg/L
    • Solve: \(60 = 100 \cdot 2^{-t/4} + 20\)
    • \(40 = 100 \cdot 2^{-t/4}\)
    • \(0.4 = 2^{-t/4}\)
    • \(-t/4 = \log_2(0.4) = -1.32\)
    • \(t = 5.28\) hours
  2. When concentration = 30:
    • \(30 = 100 \cdot 2^{-t/4} + 20\)
    • \(10 = 100 \cdot 2^{-t/4}\)
    • \(0.1 = 2^{-t/4}\)
    • \(-t/4 = \log_2(0.1) = -3.32\)
    • \(t = 13.28\) hours
  3. Cost for 100 units:
    • Cost = \(1000 + 50(100) + 0.1(100)^2\)
    • \(= 1000 + 5000 + 1000 = 7000\)
    • Scientific notation: \(7 \times 10^3\) euros
  4. Price for 5000 units demand:
    • \(5000 = 10000 \cdot 0.95^p\)
    • \(0.5 = 0.95^p\)
    • \(p = \log_{0.95}(0.5) = \frac{\ln(0.5)}{\ln(0.95)} = 13.51\)
    • Price: €13.51

Problem 4: Advanced Synthesis

Solve these challenging integration problems:

  1. If \(y = \sqrt{x + y}\) and \(y = 3\), find x.

  2. Simplify: \(\frac{\log_2(32) - \log_4(64) + \log_8(512)}{\log_{16}(256)}\)

  3. If \(a^3 + b^3 = 9\) and \(a + b = 3\), find \(ab\).

  1. Given \(y = 3\) and \(y = \sqrt{x + y}\):
    • \(3 = \sqrt{x + 3}\)
    • \(9 = x + 3\)
    • \(x = 6\)
  2. Convert to common base:
    • \(\log_2(32) = 5\)
    • \(\log_4(64) = \log_4(4^3) = 3\)
    • \(\log_8(512) = \log_8(8^3) = 3\)
    • \(\log_{16}(256) = \log_{16}(16^2) = 2\)
    • Result: \(\frac{5 - 3 + 3}{2} = \frac{5}{2}\)
  3. Use identity \(a^3 + b^3 = (a + b)^3 - 3ab(a + b)\):
    • \(9 = 3^3 - 3ab(3)\)
    • \(9 = 27 - 9ab\)
    • \(9ab = 18\)
    • \(ab = 2\)

Problem 5: Challenge - Research Project

A research institute plans a 5-year project:

Funding: Initial €1 million, growing each year by factor 1.08

  • Amount after t years = \(1000000 \cdot 1.08^t\)

Papers Published: Cumulative papers after t years = \(5 \cdot \ln(t + 1) + 2t\) (Use \(\ln(2) ≈ 0.69\), \(\ln(3) ≈ 1.10\), \(\ln(4) ≈ 1.39\))

Annual Cost: Cost in year t = \(200000 \cdot 1.05^t + 50000\sqrt{t + 1}\)

  1. What is the funding amount after 5 years? Express in scientific notation.

  2. How many papers will be published by the end of year 3?

  3. Calculate the total cost for years 0, 1, and 2.

  1. Funding after 5 years:
    • Amount = \(1000000 \times 1.08^5\)
    • \(= 1000000 \times 1.469 = 1,469,000\)
    • Scientific notation: \(1.469 \times 10^6\) euros
  2. Papers by year 3:
    • Papers = \(5\ln(4) + 2(3)\)
    • \(= 5(1.39) + 6 = 6.95 + 6 = 12.95\)
    • Approximately 13 papers
  3. Total cost for first 3 years:
    • Year 0: \(200000 + 50000(1) = \text{€}250,000\)
    • Year 1: \(200000(1.05) + 50000\sqrt{2} = 210000 + 70711 = \text{€}280,711\)
    • Year 2: \(200000(1.1025) + 50000\sqrt{3} = 220500 + 86603 = \text{€}307,103\)
    • Total: €837,814

Problem 6: Solving Linear Inequalities (x)

Solve the following inequalities and express the solution set using interval notation.

  1. \(5x + 3 \leq 18\)

  2. \(21 - 2x > 9\)

  3. \(3(x + 4) \geq 5x - 8\)

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

  1. \(5x \leq 15 \implies x \leq 3\). Interval: \((-\infty, 3]\)

  2. \(-2x > -12 \implies x < 6\) (sign flips). Interval: \((-\infty, 6)\)

  3. \(3x + 12 \geq 5x - 8 \implies 20 \geq 2x \implies 10 \geq x\). Interval: \((-\infty, 10]\)

  4. Multiply by 4: \(x - 1 < 2(x + 3) \implies x - 1 < 2x + 6 \implies -7 < x\). Interval: \((-7, \infty)\)

Problem 7: Break-Even Analysis (xx)

A company is launching a new smartwatch with the following cost structure:

  • Fixed monthly costs (rent, salaries, insurance): $12,000
  • Variable cost per watch (materials, assembly): $85
  • Planned selling price: $249

  1. Write an equation for the total cost when x watches are produced in the first month.

  2. How many watches must be sold in the first month to break even on monthly operations?

Using IDEA Method:

  1. Identify: Need total cost equation for x watches in first month
    • Develop: Total Cost = Fixed costs + Variable costs
    • Execute: Total Cost = 12,000 + 85x
    • Assess: Equation shows $12,000 base cost plus $85 per unit
  2. Identify: Find x where Revenue = Monthly Cost
    • Develop: Revenue = 249x, Total Cost = 12,000 + 85x
    • Execute:
      • 249x = 12,000 + 85x
      • 164x = 12,000
      • x = 12,000/164 ≈ 73.17
    • Assess: Need to sell 74 watches (round up for break-even!)

Problem 8: Motion and Meeting Problem (xxx)

Two delivery services are coordinating a package handoff.

  • QuickShip leaves City A heading east toward City B at 65 km/h.
  • At the same time, FastTrack leaves City B heading west toward City A at 85 km/h.
  • The cities are 375 km apart.

  1. How long until the drivers meet?

  2. How far from City A will they meet?

  3. If QuickShip had left 30 minutes earlier, where would they meet?

  1. Meeting time:
    • Let t = time in hours until meeting
    • Distance by QuickShip: 65t
    • Distance by FastTrack: 85t
    • Total distance: 65t + 85t = 375
    • 150t = 375
    • t = 2.5 hours
  2. Distance from City A:
    • QuickShip travels: 65 × 2.5 = 162.5 km from City A
    • Verification: FastTrack travels 85 × 2.5 = 212.5 km
    • Total: 162.5 + 212.5 = 375 km
  3. With 30-minute head start:
    • QuickShip travels alone for 0.5 hours: 65 × 0.5 = 32.5 km
    • Remaining distance: 375 - 32.5 = 342.5 km
    • Let t = time after FastTrack starts
    • 65t + 85t = 342.5
    • 150t = 342.5
    • t = 2.283 hours
    • Distance from City A: 32.5 + 65(2.283) = 32.5 + 148.4 = 180.9 km

Problem 9: Business Decision (xx)

A freelance designer has two pricing plans for a project:

  • Plan A: A flat fee of 1,200.
  • Plan B: An initial fee of 500 plus 35 per hour.

Let \(h\) be the number of hours the project takes.

  1. Write an expression for the total cost of each plan (e.g., Cost A, Cost B).
  2. For what number of hours are the two plans equal in cost?
  3. If the designer estimates the project will take 25 hours, which plan is cheaper for the client?

  1. Cost A = \(1200\). Cost B = \(500 + 35h\).

  2. Set costs equal: \(1200 = 500 + 35h \implies 700 = 35h \implies h = 20\). The plans cost the same at 20 hours.

  3. At 25 hours:

    • Cost A = 1200
    • Cost B = \(500 + 35(25) = 500 + 875 = 1375\) Plan A is cheaper for the client if the project takes 25 hours.

Problem 10: Investment Threshold (xx)

An investor has 20,000 to invest. They decide to put some money into a safe bond that yields 3% annual interest and the rest into a riskier stock fund that is projected to yield 8% annual interest.

What is the minimum amount of money the investor must put into the stock fund to ensure a total annual return of at least 1,000?

Let \(s\) be the amount invested in the stock fund. Then, \(20000 - s\) is the amount invested in the bond.

The total return is the sum of the returns from each investment. Total Return = (Return from Stocks) + (Return from Bonds) Total Return = \(0.08s + 0.03(20000 - s)\)

The goal is for the total return to be at least 1,000. \(0.08s + 0.03(20000 - s) \geq 1000\)

Now, solve the inequality: \(0.08s + 600 - 0.03s \geq 1000\) \(0.05s + 600 \geq 1000\) \(0.05s \geq 400\) \(s \geq \frac{400}{0.05}\) \(s \geq 8000\)

The minimum amount the investor must put into the stock fund is 8,000.