Tasks 01-06 - Synthesis & Problem Solving

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: \(200000 + 50000(1) = €250,000\)
    • Year 1: \(200000(1.05) + 50000\sqrt{2} = 210000 + 70711 = €280,711\)
    • Year 2: \(200000(1.1025) + 50000\sqrt{3} = 220500 + 86603 = €307,103\)
    • Total: €837,814