Tasks 02-05 - Exponential, Logarithmic & Complex Word Problems

Section 02: Equations & Problem-Solving Strategies

Instructions

Complete these problems to master advanced exponential and logarithmic equations, as well as complex multi-step word problems. These problems integrate all equation-solving techniques from Section 02.

Problem 1: Exponential Equations (x)

Solve each exponential equation:

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

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

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

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

  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: \(4^x = (2^2)^x = (2^x)^2\)
    • 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: \(x = 2\)

Problem 2: Logarithmic Equations (x)

Solve each logarithmic equation, stating domain restrictions:

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

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

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

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

  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: \(x > 0\)
    • 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: \(x > 0, x \neq 1\)
    • Change base: \(\log_8(x) = \frac{\log_2(x)}{3}\)
    • 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: \(x > 0, x \neq 1\)
    • Convert: \(x^4 = 16\)
    • So: \(x^4 = 2^4\)
    • Solution: \(x = 2\) (also \(x = -2\), but rejected due to domain)

Problem 3: Mixed Exponential-Logarithmic (xx)

Solve these equations that involve both exponential and logarithmic expressions:

  1. \(3^{\log_3(x)} + x = 10\)

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

Note, there might be multiple solutions or no solution at all!

  1. \(3^{\log_3(x)} + x = 10\)
    • Note: \(3^{\log_3(x)} = x\) (for \(x > 0\))
    • So: \(x + x = 10\)
    • Solution: \(x = 5\)
    • Check: \(3^{\log_3(5)} + 5 = 5 + 5 = 10\) ✓
  2. \(\log_2(2^x + 1) = x - 1\)
    • Convert: \(2^x + 1 = 2^{x-1}\)
    • Multiply by 2: \(2 \cdot 2^x + 2 = 2^x\)
    • This gives: \(2^x + 2 = 0\)
    • Since \(2^x > 0\), this is impossible
    • No solution

Problem 4: Growth and Decay Applications (xx)

A radioactive substance decays according to \(A(t) = A_0 e^{-kt}\) where \(t\) is in years.

  1. If 70% remains after 5 years, find the decay constant \(k\)
  2. Find the half-life of the substance
  3. How long until only 10% remains?
  4. If we start with 100 grams, when will exactly 25 grams remain?
  1. Find decay constant:
    • Given: \(0.7A_0 = A_0 e^{-5k}\)
    • Simplify: \(0.7 = e^{-5k}\)
    • Take ln: \(\ln(0.7) = -5k\)
    • Solve: \(k = -\frac{\ln(0.7)}{5} = \frac{\ln(10/7)}{5} \approx 0.0713\) per year
  2. Find half-life:
    • Half-life when \(A(t) = 0.5A_0\)
    • So: \(0.5 = e^{-kt}\)
    • \(\ln(0.5) = -kt\)
    • \(t = -\frac{\ln(0.5)}{k} = \frac{\ln(2)}{0.0713} \approx 9.72\) years
  3. Time for 10% remaining:
    • Need: \(0.1A_0 = A_0 e^{-kt}\)
    • So: \(0.1 = e^{-kt}\)
    • \(\ln(0.1) = -kt\)
    • \(t = -\frac{\ln(0.1)}{k} = \frac{\ln(10)}{0.0713} \approx 32.3\) years
  4. 25 grams from 100 grams:
    • Need: \(25 = 100e^{-kt}\)
    • So: \(0.25 = e^{-kt}\)
    • \(\ln(0.25) = -kt\)
    • \(t = -\frac{\ln(0.25)}{k} = \frac{\ln(4)}{0.0713} \approx 19.4\) years

Problem 5: Investment Comparison (xx)

An investor has €30,000 to invest. She’s considering three options:

  • Option A: 6% annual interest, compounded yearly
  • Option B: 5.8% annual interest, compounded monthly
  • Option C: 5.7% annual interest, compounded continuously
  1. Write the formula for each option after \(t\) years
  2. Which option yields the most after 10 years?
  3. If she wants €50,000, how long would each option take?
  1. Formulas:
    • Option A: \(A(t) = 30000(1.06)^t\)
    • Option B: \(B(t) = 30000(1 + \frac{0.058}{12})^{12t} = 30000(1.00483)^{12t}\)
    • Option C: \(C(t) = 30000e^{0.057t}\)
  2. After 10 years:
    • Option A: \(A(10) = 30000(1.06)^{10} \approx €53,725\)
    • Option B: \(B(10) = 30000(1.00483)^{120} \approx €53,456\)
    • Option C: \(C(10) = 30000e^{0.57} \approx €53,049\)
    • Option A yields the most
  3. Time to reach €50,000:
    • Option A: \(50000 = 30000(1.06)^t\)
      • \((1.06)^t = \frac{5}{3}\)
      • \(t = \frac{\ln(5/3)}{\ln(1.06)} \approx 8.77\) years
    • Option B: \(50000 = 30000(1.00483)^{12t}\)
      • \((1.00483)^{12t} = \frac{5}{3}\)
      • \(t = \frac{\ln(5/3)}{12\ln(1.00483)} \approx 8.82\) years
    • Option C: \(50000 = 30000e^{0.057t}\)
      • \(e^{0.057t} = \frac{5}{3}\)
      • \(t = \frac{\ln(5/3)}{0.057} \approx 8.96\) years

Problem 6: Population Dynamics (xxx)

Two bacterial cultures are growing in a lab. Culture A starts with 500 bacteria and doubles every 3 hours. Culture B starts with 800 bacteria and grows according to \(B(t) = 800e^{0.15t}\) where \(t\) is in hours.

  1. Write the growth equation for Culture A
  2. When will the populations be equal?
  1. Culture A equation:
    • Doubles every 3 hours: \(A(t) = 500 \cdot 2^{t/3}\)
  2. When populations are equal:
    • Set equal: \(500 \cdot 2^{t/3} = 800e^{0.15t}\)
    • Divide by 500: \(2^{t/3} = 1.6e^{0.15t}\)
    • Take ln: \(\frac{t}{3}\ln(2) = \ln(1.6) + 0.15t\)
    • \(0.231t = 0.470 + 0.15t\)
    • \(0.081t = 0.470\)
    • \(t \approx 5.8\) hours