Tasks 01-05 - Logarithms & Substitution

Advanced Algebraic Tools

Problem 1: Substitution for Factorization

Use substitution to factor the following expressions completely:

\(x^4 + 5x^2 + 6\)
\(4x^4 - 13x^2 + 9\)
\(x^6 - 9x^3 + 8\)
\((x^2 + 3x)^2 - 2(x^2 + 3x) - 24\)
\(16x^4 - 8x^2 + 1\)

Solution

\(x^4 + 5x^2 + 6\)
- Let \(u = x^2\): \(u^2 + 5u + 6 = (u + 2)(u + 3)\)
- Substitute back: \((x^2 + 2)(x^2 + 3)\)
\(4x^4 - 13x^2 + 9\)
- Let \(u = x^2\): \(4u^2 - 13u + 9 = (4u - 9)(u - 1)\)
- Substitute back: \((4x^2 - 9)(x^2 - 1) = (2x + 3)(2x - 3)(x + 1)(x - 1)\)
\(x^6 - 9x^3 + 8\)
- Let \(u = x^3\): \(u^2 - 9u + 8 = (u - 1)(u - 8)\)
- Substitute back: \((x^3 - 1)(x^3 - 8)\)
- Factor further: \((x - 1)(x^2 + x + 1)(x - 2)(x^2 + 2x + 4)\)
\((x^2 + 3x)^2 - 2(x^2 + 3x) - 24\)
- Let \(u = x^2 + 3x\): \(u^2 - 2u - 24 = (u - 6)(u + 4)\)
- Substitute back: \((x^2 + 3x - 6)(x^2 + 3x + 4)\)
\(16x^4 - 8x^2 + 1\)
- Let \(u = x^2\): \(16u^2 - 8u + 1 = (4u - 1)^2\)
- Substitute back: \((4x^2 - 1)^2 = [(2x + 1)(2x - 1)]^2 = (2x + 1)^2(2x - 1)^2\)

Evaluate the following without a calculator:

Solution

\(\log_2(64) = \log_2(2^6) = 6\)
\(\log_3(\frac{1}{27}) = \log_3(3^{-3}) = -3\)
\(\log_5(125) = \log_5(5^3) = 3\)
\(\log_{10}(0.001) = \log_{10}(10^{-3}) = -3\)
\(\log_4(8)\): Since \(4 = 2^2\) and \(8 = 2^3\)
- Let \(\log_4(8) = x\), then \(4^x = 8\)
- \((2^2)^x = 2^3\)
- \(2^{2x} = 2^3\)
- \(2x = 3\), so \(x = \frac{3}{2}\)
\(\ln(e^3) = 3\)

Simplify the following expressions using logarithm laws:

\(\log_2(16) + \log_2(8) - \log_2(4)\)
\(\log(50) + \log(20) - \log(100)\)
\(3\log_5(x) - \log_5(x^2) + \log_5(25)\)
\(\log_3(81) - 2\log_3(9) + \log_3(27)\)
Express as a single logarithm: \(\frac{1}{2}\ln(x) + 3\ln(y) - \ln(z)\)

Solution

\(\log_2(16) + \log_2(8) - \log_2(4)\)
- Direct: \(4 + 3 - 2 = 5\)
- Or: \(\log_2(\frac{16 \times 8}{4}) = \log_2(32) = 5\)
\(\log(50) + \log(20) - \log(100)\)
- \(= \log(\frac{50 \times 20}{100}) = \log(10) = 1\)
\(3\log_5(x) - \log_5(x^2) + \log_5(25)\)
- \(= \log_5(x^3) - \log_5(x^2) + \log_5(25)\)
- \(= \log_5(\frac{x^3 \times 25}{x^2}) = \log_5(25x) = \log_5(25) + \log_5(x)\)
- \(= 2 + \log_5(x)\)
\(\log_3(81) - 2\log_3(9) + \log_3(27)\)
- \(= 4 - 2(2) + 3 = 4 - 4 + 3 = 3\)
\(\frac{1}{2}\ln(x) + 3\ln(y) - \ln(z)\)
- \(= \ln(x^{1/2}) + \ln(y^3) - \ln(z)\)
- \(= \ln(\frac{\sqrt{x} \cdot y^3}{z})\)

Solve the following equations:

Solution: Logarithmic Equations

\(\log_3(x + 5) = 2\)
- \(x + 5 = 3^2 = 9\)
- \(x = 4\)
\(\log(2x - 1) - \log(x + 2) = 0\)
- \(\log(\frac{2x - 1}{x + 2}) = 0\)
- \(\frac{2x - 1}{x + 2} = 10^0 = 1\)
- \(2x - 1 = x + 2\)
- \(x = 3\)
\(2^{\log_2(x)} = 8\)
- By definition: \(2^{\log_2(x)} = x\)
- So \(x = 8\)
\(\log_x(49) = 2\)
- \(x^2 = 49\)
- \(x = 7\) (x must be positive and ≠ 1)

Solution: Binomial Expansion

\((2x - 3)^4\)
- Coefficients: 1, 4, 6, 4, 1
- \(= (2x)^4 + 4(2x)^3(-3) + 6(2x)^2(-3)^2 + 4(2x)(-3)^3 + (-3)^4\)
- \(= 16x^4 - 96x^3 + 216x^2 - 216x + 81\)
\((3x + 1)^6\) Using the binomial theorem: \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)

With \(a = 3x\), \(b = 1\), and \(n = 6\):
- \(k = 0\): \(\binom{6}{0}(3x)^6(1)^0 = 1 \times 729x^6 = 729x^6\)
- \(k = 1\): \(\binom{6}{1}(3x)^5(1)^1 = 6 \times 243x^5 = 1458x^5\)
- \(k = 2\): \(\binom{6}{2}(3x)^4(1)^2 = 15 \times 81x^4 = 1215x^4\)
- \(k = 3\): \(\binom{6}{3}(3x)^3(1)^3 = 20 \times 27x^3 = 540x^3\)
- \(k = 4\): \(\binom{6}{4}(3x)^2(1)^4 = 15 \times 9x^2 = 135x^2\)
- \(k = 5\): \(\binom{6}{5}(3x)^1(1)^5 = 6 \times 3x = 18x\)
- \(k = 6\): \(\binom{6}{6}(3x)^0(1)^6 = 1 \times 1 = 1\)
\(= 729x^6 + 1458x^5 + 1215x^4 + 540x^3 + 135x^2 + 18x + 1\)

An investment grows from €5,000 to €8,000 in 6 years with continuous compounding. Find the interest rate.

Solution: Applications