Tasks 02-03 - Fractional, Radical, Exponential & Logarithmic Equations

Section 02: Equations & Problem-Solving Strategies

Instructions

Complete these problems to master solving fractional, radical, and cubic equations. Pay special attention to domain restrictions and checking for extraneous solutions.

Problem 1: Domain Restrictions (x)

For each rational equation, identify the domain restrictions BEFORE solving:

  1. \(\frac{5}{x-3} = 2\)

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

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

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

  5. \(\frac{x+1}{x-3} = \frac{x-2}{x+3}\)

  1. \(\frac{5}{x-3} = 2\)
    • Domain restriction: \(x \neq 3\)
    • Solve: \(5 = 2(x-3)\)
    • \(5 = 2x - 6\)
    • \(11 = 2x\)
    • \(x = 5.5\) (valid, since \(5.5 \neq 3\))
  2. \(\frac{x}{x+2} + \frac{3}{x-1} = 1\)
    • Domain restrictions: \(x \neq -2, x \neq 1\)
    • LCD: \((x+2)(x-1)\)
    • \(x(x-1) + 3(x+2) = (x+2)(x-1)\)
    • \(x^2 - x + 3x + 6 = x^2 + x - 2\)
    • \(2x + 6 = x - 2\)
    • \(x = -8\) (valid)
  3. \(\frac{2x}{x^2-4} = \frac{1}{x-2}\)
    • Factor denominator: \(x^2 - 4 = (x-2)(x+2)\)
    • Domain restrictions: \(x \neq 2, x \neq -2\)
    • Simplify: \(\frac{2x}{(x-2)(x+2)} = \frac{1}{x-2}\)
    • Cross multiply: \(2x = x + 2\)
    • \(x = 2\) (INVALID - violates domain!)
    • No solution
  4. \(\frac{1}{x} + \frac{1}{x^2} = \frac{1}{2}\)
    • Domain restriction: \(x \neq 0\)
    • LCD: \(2x^2\)
    • \(2x + 2 = x^2\)
    • \(x^2 - 2x - 2 = 0\)
    • \(x = \frac{2 \pm \sqrt{4+8}}{2} = 1 \pm \sqrt{3}\)
    • Both solutions valid (neither equals 0)
  5. \(\frac{x+1}{x-3} = \frac{x-2}{x+3}\)
    • Domain restrictions: \(x \neq 3, x \neq -3\)
    • Cross multiply: \((x+1)(x+3) = (x-2)(x-3)\)
    • \(x^2 + 4x + 3 = x^2 - 5x + 6\)
    • \(9x = 3\)
    • \(x = \frac{1}{3}\) (valid)

Problem 2: Solving Rational Equations (x)

Solve each equation, checking domain restrictions:

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

  2. \(\frac{x+2}{x-1} - \frac{x-1}{x+2} = 0\)

  3. \(\frac{2}{x-3} + \frac{x}{x+3} = \frac{12}{x^2-9}\)

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

  1. \(\frac{4}{x} - \frac{3}{2x} = \frac{1}{4}\)
    • Domain: \(x \neq 0\)
    • LCD: \(4x\)
    • \(16 - 6 = x\)
    • \(x = 10\) (valid)
  2. \(\frac{x+2}{x-1} - \frac{x-1}{x+2} = 0\)
    • Domain: \(x \neq 1, x \neq -2\)
    • \(\frac{x+2}{x-1} = \frac{x-1}{x+2}\)
    • Cross multiply: \((x+2)^2 = (x-1)^2\)
    • \(x^2 + 4x + 4 = x^2 - 2x + 1\)
    • \(6x = -3\)
    • \(x = -\frac{1}{2}\) (valid)
  3. \(\frac{2}{x-3} + \frac{x}{x+3} = \frac{12}{x^2-9}\)
    • Note: \(x^2 - 9 = (x-3)(x+3)\)
    • Domain: \(x \neq 3, x \neq -3\)
    • LCD: \((x-3)(x+3)\)
    • \(2(x+3) + x(x-3) = 12\)
    • \(2x + 6 + x^2 - 3x = 12\)
    • \(x^2 - x - 6 = 0\)
    • \((x-3)(x+2) = 0\)
    • \(x = 3\) (invalid) or \(x = -2\) (valid)
    • Solution: \(x = -2\)
  4. \(\frac{1}{x-2} + \frac{2}{x+1} = \frac{3}{x-2}\)
    • Domain: \(x \neq 2, x \neq -1\)
    • Simplify: \(\frac{2}{x+1} = \frac{3-1}{x-2} = \frac{2}{x-2}\)
    • Therefore: \(x + 1 = x - 2\)
    • \(1 = -2\) (impossible)
    • No solution

Problem 3: Basic Radical Equations (x)

Solve each radical equation and check for extraneous solutions:

  1. \(\sqrt{x + 5} = 3\)

  2. \(\sqrt{2x - 1} = x - 2\)

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

  4. \(x = \sqrt{x + 6}\)

  5. \(\sqrt{4x + 1} = 2x - 1\)

  1. \(\sqrt{x + 5} = 3\)
    • Square: \(x + 5 = 9\)
    • \(x = 4\)
    • Check: \(\sqrt{4 + 5} = \sqrt{9} = 3\) ✓
  2. \(\sqrt{2x - 1} = x - 2\)
    • Square: \(2x - 1 = x^2 - 4x + 4\)
    • \(x^2 - 6x + 5 = 0\)
    • \((x - 5)(x - 1) = 0\)
    • \(x = 5\) or \(x = 1\)
    • Check \(x = 5\): \(\sqrt{9} = 3\) and \(5 - 2 = 3\) ✓
    • Check \(x = 1\): \(\sqrt{1} = 1\) but \(1 - 2 = -1\) ✗
    • Solution: \(x = 5\)
  3. \(\sqrt{x^2 + 3} = 2\)
    • Square: \(x^2 + 3 = 4\)
    • \(x^2 = 1\)
    • \(x = \pm 1\)
    • Check both: \(\sqrt{1 + 3} = 2\) ✓
    • Solutions: \(x = 1\) or \(x = -1\)
  4. \(x = \sqrt{x + 6}\)
    • Square: \(x^2 = x + 6\)
    • \(x^2 - x - 6 = 0\)
    • \((x - 3)(x + 2) = 0\)
    • \(x = 3\) or \(x = -2\)
    • Check \(x = 3\): \(3 = \sqrt{9} = 3\) ✓
    • Check \(x = -2\): \(-2 \neq \sqrt{4} = 2\) ✗
    • Solution: \(x = 3\)
  5. \(\sqrt{4x + 1} = 2x - 1\)
    • Square: \(4x + 1 = 4x^2 - 4x + 1\)
    • \(4x^2 - 8x = 0\)
    • \(4x(x - 2) = 0\)
    • \(x = 0\) or \(x = 2\)
    • Check \(x = 0\): \(\sqrt{1} = 1\) but \(-1 \neq 1\) ✗
    • Check \(x = 2\): \(\sqrt{9} = 3\) and \(4 - 1 = 3\) ✓
    • Solution: \(x = 2\)

Problem 4: Multiple Radicals (xx)

Solve equations with multiple radical terms:

  1. \(\sqrt{x + 3} + \sqrt{x} = 3\)

  2. \(\sqrt{2x + 5} - \sqrt{x + 2} = 1\)

  3. \(\sqrt{x + 8} = 2 + \sqrt{x}\)

  1. \(\sqrt{x + 3} + \sqrt{x} = 3\)
    • Isolate: \(\sqrt{x + 3} = 3 - \sqrt{x}\)
    • Square: \(x + 3 = 9 - 6\sqrt{x} + x\)
    • \(3 = 9 - 6\sqrt{x}\)
    • \(6\sqrt{x} = 6\)
    • \(\sqrt{x} = 1\), so \(x = 1\)
    • Check: \(\sqrt{4} + \sqrt{1} = 2 + 1 = 3\) ✓
  2. \(\sqrt{2x + 5} - \sqrt{x + 2} = 1\)
    • Isolate: \(\sqrt{2x + 5} = 1 + \sqrt{x + 2}\)
    • Square: \(2x + 5 = 1 + 2\sqrt{x + 2} + x + 2\)
    • \(x + 2 = 2\sqrt{x + 2}\)
    • \(\frac{x + 2}{2} = \sqrt{x + 2}\)
    • Square again: \(\frac{(x + 2)^2}{4} = x + 2\)
    • \((x + 2)^2 = 4(x + 2)\)
    • \((x + 2)(x + 2 - 4) = 0\)
    • \((x + 2)(x - 2) = 0\)
    • \(x = -2\) (makes \(\sqrt{x + 2} = 0\)) or \(x = 2\)
    • Check \(x = 2\): \(\sqrt{9} - \sqrt{4} = 3 - 2 = 1\) ✓
    • Check \(x = -2\): \(\sqrt{1} - 0 = 1\) ✓
    • Solutions: \(x = -2\) or \(x = 2\)
  3. \(\sqrt{x + 8} = 2 + \sqrt{x}\)
    • Square: \(x + 8 = 4 + 4\sqrt{x} + x\)
    • \(4 = 4\sqrt{x}\)
    • \(\sqrt{x} = 1\)
    • \(x = 1\)
    • Check: \(\sqrt{9} = 3\) and \(2 + 1 = 3\) ✓

Problem 5: Cubic Factoring (xx)

Factor and solve each cubic equation:

  1. \(x^3 + x^2 - 6x = 0\)

  2. \(x^3 - 3x^2 - 4x + 12 = 0\)

  3. \(x^3 + 3x^2 - 13x - 15 = 0\)

  1. \(x^3 + x^2 - 6x = 0\)
    • Factor out x: \(x(x^2 + x - 6) = 0\)
    • \(x(x + 3)(x - 2) = 0\)
    • Solutions: \(x = 0, -3, 2\)
  2. \(x^3 - 3x^2 - 4x + 12 = 0\)
    • Group: \(x^2(x - 3) - 4(x - 3) = 0\)
    • \((x - 3)(x^2 - 4) = 0\)
    • \((x - 3)(x - 2)(x + 2) = 0\)
    • Solutions: \(x = -2, 2, 3\)
  3. \(x^3 + 3x^2 - 13x - 15 = 0\)
    • Try: Test \(x = 3\)
    • \(27 + 27 - 39 - 15 = 0\) ✓
    • Factor: \((x - 3)(x^2 + 6x + 5) = 0\)
    • \((x - 3)(x + 5)(x + 1) = 0\)
    • Solutions: \(x = -5, -1, 3\)

Problem 6: Work Rate Problem (xx)

Two pipes can fill a tank together in 4 hours. The larger pipe alone can fill it 3 hours faster than the smaller pipe alone.

  1. Set up the equation with appropriate variables
  2. Identify domain restrictions
  3. Solve for the individual filling times
  4. Verify your answer makes practical sense
  1. Setup:
    • Let \(x\) = hours for smaller pipe alone
    • Then \(x - 3\) = hours for larger pipe alone
    • Smaller pipe rate: \(\frac{1}{x}\) tanks/hour
    • Larger pipe rate: \(\frac{1}{x-3}\) tanks/hour
    • Combined rate: \(\frac{1}{4}\) tanks/hour
    • Equation: \(\frac{1}{x} + \frac{1}{x-3} = \frac{1}{4}\)
  2. Domain restrictions:
    • \(x \neq 0\) and \(x \neq 3\)
    • Also need \(x > 3\) for practical sense (larger pipe is faster)
  3. Solve:
    • LCD: \(4x(x-3)\)
    • \(4(x-3) + 4x = x(x-3)\)
    • \(4x - 12 + 4x = x^2 - 3x\)
    • \(x^2 - 11x + 12 = 0\)
    • Using quadratic formula: \(x = \frac{11 \pm \sqrt{121 - 48}}{2} = \frac{11 \pm \sqrt{73}}{2}\)
    • \(x \approx 9.77\) or \(x \approx 1.23\)
    • Since we need \(x > 3\): \(x \approx 9.77\) hours
  4. Verify:
    • Smaller pipe: 9.77 hours alone
    • Larger pipe: 6.77 hours alone
    • Combined rate: \(\frac{1}{9.77} + \frac{1}{6.77} \approx 0.102 + 0.148 = 0.25 = \frac{1}{4}\) ✓

Problem 7: Investment Returns (xxx)

An investor divides €10,000 between two funds. Fund A returns interest at rate \(r\)%, while Fund B returns at rate \((r+2)\)%. After one year, the total interest earned is €650, if the amount in Fund B is €2,000 more than in Fund A:

  1. Set up equations for the investment amounts and returns
  2. Find the interest rates for both funds
  3. Calculate the exact amounts invested in each fund
  4. What would the total return be if all money was in Fund B?
  1. Setup:
    • Let \(x\) = amount in Fund A
    • Then \(x + 2000\) = amount in Fund B
    • Total: \(x + (x + 2000) = 10000\), so \(x = 4000\)
    • Fund A has €4,000, Fund B has €6,000
    • Interest equation: \(\frac{r}{100}(4000) + \frac{r+2}{100}(6000) = 650\)
  2. Find rates:
    • \(40r + 60(r + 2) = 650\)
    • \(40r + 60r + 120 = 650\)
    • \(100r = 530\)
    • \(r = 5.3\%\)
    • Fund A rate: 5.3%
    • Fund B rate: 7.3%
  3. Verify amounts:
    • Fund A: €4,000 at 5.3% = €212
    • Fund B: €6,000 at 7.3% = €438
    • Total interest: €212 + €438 = €650 ✓
  4. All in Fund B:
    • €10,000 at 7.3% = €730
    • Additional return: €730 - €650 = €80 more

Problem 8: Calculator – Cubic Equations (xx)

Solve using the polynomial solver:

  1. \(x^3 - 6x^2 + 11x - 6 = 0\)

  2. \(x^3 - 3x^2 - 4x + 12 = 0\)

  1. \(x^3 - 6x^2 + 11x - 6 = 0\)
    • Coefficients: a = 1, b = -6, c = 11, d = -6
    • Solutions: \(x_1 = 1\), \(x_2 = 2\), \(x_3 = 3\)
    • Verification: \((x-1)(x-2)(x-3) = x^3 - 6x^2 + 11x - 6\) ✓
  2. \(x^3 - 3x^2 - 4x + 12 = 0\)
    • Coefficients: a = 1, b = -3, c = -4, d = 12
    • Solutions: \(x_1 = -2\), \(x_2 = 2\), \(x_3 = 3\)

Problem 9: Calculator – SOLVE Function - Newton’s Method (xx)

Use the SOLVE function (SHIFT + CALC) to find solutions numerically:

  1. Find one solution to \(x^3 - 5x + 3 = 0\) (start with initial guess x = 1)

  2. Solve \(\sin(x) = 0.5\) for x in radians (start with x = 0.5)

  3. Find where the derivative of \(f(x) = x^3 - 6x\) equals zero

  1. \(x^3 - 5x + 3 = 0\)
    • Enter equation, press SHIFT CALC
    • Initial guess: 1
    • Answer: \(x \approx 0.6566\) (one of three roots)
    • Try starting with x = -3 to find another root: \(x \approx -2.331\)
  2. \(\sin(x) = 0.5\)
    • Set angle mode to Radians (R)
    • Enter: sin(x) = 0.5
    • Initial guess: 0.5
    • Answer: \(x \approx 0.5236\) (which is \(\frac{\pi}{6}\))
  3. Where \(f'(x) = 0\) for \(f(x) = x^3 - 6x\)
    • \(f'(x) = 3x^2 - 6 = 0\)
    • Enter: 3x² - 6 = 0 or use d/dx template
    • Answers: \(x = \pm\sqrt{2} \approx \pm 1.414\)

Problem 10: 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 11: 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 12: 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 13: 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