Session 01-05 - Logarithms & Substitution

Section 01: Mathematical Foundations & Algebra

Author

Dr. Nikolai Heinrichs & Dr. Tobias Vlćek

Entry Quiz

Quick Review from Last Session

Complete individually, then we discuss

Factor completely: \(x^3 - 27\)
Simplify: \(\sqrt{48} + \sqrt{12} - \sqrt{75}\)
Rationalize: \(\frac{3}{\sqrt{5} - 2}\)
Factor using AC method: \(2x^2 + 7x + 3\)

. . .

Let’s review together!

Student Presentations

Homework Showcase

20 minutes for presentations and discussion

Present and discuss your solutions from Tasks 01-04
Share any challenging problems or interesting approaches
Use this time to clarify concepts before we move forward

. . .

Today we build on factorization and radicals with new powerful tools!

Substitution for Factorization

What is Substitution for Factorization?

Making complex expressions simpler by introducing a new variable

Sometimes factorization becomes easier when we substitute part of an expression with a simpler variable.

Strategy: Replace a repeated expression with a single variable
Factor the simpler expression
Substitute back to get the final answer
Why it works: Reduces cognitive load and reveals hidden patterns

. . .

Look for expressions that appear multiple times or have a clear “inner” structure!

When to Use Substitution

Recognize these common patterns

Quadratic in form: \((x^2)^2 + 5(x^2) + 6\)
Repeated expressions: \((2x + 1)^2 - 3(2x + 1) - 10\)
Complex nested terms: \(\sqrt{x + 1} - 2\sqrt{x + 1} + 1\)
Trigonometric expressions: \(\sin^2(x) + 3\sin(x) + 2\)

. . .

The key is identifying what to substitute - look for the “building block” that repeats!

Substitution Examples

Let’s work through some examples step by step

Factor \(x^4 - 13x^2 + 36\)

Step 1: Let \(u = x^2\), so \(x^4 = u^2\)
Step 2: Substitute: \(u^2 - 13u + 36\)
Step 3: Factor: \((u - 4)(u - 9)\)
Step 4: Substitute back: \((x^2 - 4)(x^2 - 9)\)
Step 5: Factor completely: \((x - 2)(x + 2)(x - 3)(x + 3)\)

Factor \(x + 6\sqrt{x} + 8\)

Step 1: Notice this involves \(x\) and \(\sqrt{x}\), where \(x = (\sqrt{x})^2\)
Step 2: Let \(u = \sqrt{x}\), so \(x = u^2\)
Step 3: Substitute: \(u^2 + 6u + 8\)
Step 4: Factor: \((u + 2)(u + 4)\)
Step 5: Substitute back: \((\sqrt{x} + 2)(\sqrt{x} + 4)\)

Factor \(3x^6 - 11x^3 - 20\)

Step 1: Let \(u = x^3\), so \(x^6 = u^2\) and we have \(3u^2 - 11u - 20\)
Step 2: Use AC method: \(ac = 3(-20) = -60\)
Step 3: Find factors of -60 that sum to -11: (4, -15)
Step 4: Rewrite: \(3u^2 + 4u - 15u - 20\)
Step 5: Group: \(u(3u + 4) - 5(3u + 4) = (u - 5)(3u + 4)\)
Step 6: Substitute back: \((x^3 - 5)(3x^3 + 4)\)

Common Substitutions

Simplification tricks

When you see:

\(3^{2x}\) → Let \(u = 3^x\), then \(3^{2x} = u^2\)
\(\sqrt{x}\) appearing multiple times → Let \(u = \sqrt{x}\)
Symmetric expressions → Look for factoring patterns
Repeating decimals → Use algebraic method to find fraction

Practice with Substitution

Try these on your own

Work individually, then we’ll discuss solutions:

Factor: \(x^6 + 8x^3 + 16\)
Factor: \((\sqrt{x} - 2)^2 - 5(\sqrt{x} - 2) + 6\)
Factor: \(16x^4 - 81\)
Factor: \((x^2 + 3x)^2 - 8(x^2 + 3x) + 15\)

. . .

Always check if you can factor further after substituting back!

More Advanced Substitution

Sometimes you need to think a little bit more

Example: Factor \(x^{2/3} - 5x^{1/3} + 6\)

Observation: This involves fractional exponents
Key insight: Let \(u = x^{1/3}\), so \(x^{2/3} = (x^{1/3})^2 = u^2\)
Step 1: Substitute: \(u^2 - 5u + 6\)
Step 2: Factor: \((u - 2)(u - 3)\)
Step 3: Substitute back: \((x^{1/3} - 2)(x^{1/3} - 3)\)
Step 4: Can also write as: \((\sqrt[3]{x} - 2)(\sqrt[3]{x} - 3)\)

Break - 10 Minutes

Logarithms - The Basics

What is a Logarithm?

The logarithm is the inverse of exponentiation

\[\text{If } a^x = b \text{, then } \log_a(b) = x\]

Think of it as: “What power do I raise \(a\) to get \(b\)?”

\(2^3 = 8\) means \(\log_2(8) = 3\)
\(5^x = 125\) means \(x = \log_5(125) = 3\)

. . .

Standard notation:

\(\log\) without a base means \(\log_{10}\) (common logarithm)
\(\ln\) means \(\log_e\) where \(e \approx 2.718\) (natural logarithm)

Key Logarithm Properties

These follow directly from exponent laws!

Property	Formula	Why it works
\(\log_a(1) = 0\)	Because \(a^0 = 1\)	Any base to the 0 is 1
\(\log_a(a) = 1\)	Because \(a^1 = a\)	Base to the 1st is itself
\(\log_a(a^x) = x\)	Direct from definition	Inverse operations
\(a^{\log_a(x)} = x\)	Direct from definition	Inverse operations

Important: Logarithms are transcendental functions - they cannot be expressed using only algebraic operations (unlike polynomials, radicals, and rational functions).

Laws of Logarithms

These transform complex operations into simple ones

Rule	Formula	Example
Product	\(\log_a(xy) = \log_a(x) + \log_a(y)\)	\(\log(20) = \log(4) + \log(5)\)
Quotient	\(\log_a(\frac{x}{y}) = \log_a(x) - \log_a(y)\)	\(\log(\frac{100}{4}) = \log(100) - \log(4)\)
Power	\(\log_a(x^n) = n\log_a(x)\)	\(\log(8^3) = 3\log(8)\)

. . .

Common Mistake

\(\log(x + y) \neq \log(x) + \log(y)\)

There’s NO simple rule for \(\log(x + y)\)!

Working with Logarithms

Evaluating Logarithms

Find \(\log_3(81)\)

Ask: “3 to what power equals 81?”
\(3^1 = 3\), \(3^2 = 9\), \(3^3 = 27\), \(3^4 = 81\)
Therefore: \(\log_3(81) = 4\)

Simplify \(\log_2(32) + \log_2(8) - \log_2(4)\)

Method 1: Evaluate each
- \(\log_2(32) = 5\), \(\log_2(8) = 3\), \(\log_2(4) = 2\)
- Result: \(5 + 3 - 2 = 6\)
Method 2: Use laws
- \(= \log_2(\frac{32 \times 8}{4}) = \log_2(64) = 6\)

Change of Base Formula

Convert between different bases

\[\log_a(x) = \frac{\log_b(x)}{\log_b(a)} = \frac{\ln(x)}{\ln(a)} = \frac{\log(x)}{\log(a)}\]

. . .

Example: Find \(\log_5(30)\)

\(\log_5(30) = \frac{\ln(30)}{\ln(5)} = \frac{3.401}{1.609} \approx 2.113\)
Check: \(5^{2.113} \approx 30\) ✓

Individual Exercise 01

Practice logarithm skills

Evaluate: \(\log_4(64)\)
Simplify: \(\log_3(9) + \log_3(27)\)
Solve: \(\log_5(x + 4) = 2\)
Express as a single logarithm: \(2\log(x) - \log(y) + \log(3)\)

Logarithms in the Real World

Why Logarithms Matter

From protecting your hearing to predicting disasters

The Challenge: Natural phenomena span enormous ranges
Human perception: We sense changes proportionally, not linearly
The Solution: Logarithmic scales compress huge ranges into manageable numbers
Real Impact: These scales help save lives and advance science

. . .

Historical Note: Logarithms were invented in 1614 by John Napier to simplify astronomical calculations. Today, they’re essential for measuring everything from sound to earthquakes!

Scientific Applications

Why we need the decibel scale: Sound intensity ranges from \(10^{-12}\) to \(10^{12}\) watts/m² - that’s 24 orders of magnitude!

Decibel formula: \(L = 10\log(\frac{I}{I_0})\) dB

Whisper: 30 dB (1,000× threshold)
Normal conversation: 60 dB (1,000,000× threshold)
Rock concert: 110 dB (100,000,000,000× threshold)
Jet engine: 140 dB (causes immediate hearing damage!)

Health Alert: Each 10 dB increase = 10× intensity. That rock concert isn’t just “a bit louder” - it’s 1,000× more intense than conversation!

The problem with linear scales: Earthquake energy ranges from equivalents of small explosions to thousands of atomic bombs!

Richter scale: \(M = \log_{10}(\frac{A}{A_0})\)

Magnitude 3: Barely felt (like a large truck passing)
Magnitude 5: Light damage (100× stronger than Mag 3)
Magnitude 7: Major earthquake (10,000× stronger than Mag 3)
Magnitude 9: Great earthquake (1,000,000× stronger than Mag 3)

Financial Applications

Why Natural Logarithm for Finance?

The connection to continuous growth

Any logarithm works: \(t = \frac{\log_{10}(2)}{\log_{10}(1 + r)} = \frac{\ln(2)}{\ln(1 + r)}\)
But ln is natural because it connects to continuous compounding
Continuous compounding formula: \(A = Pe^{rt}\) (where \(e \approx 2.718\))
Why e appears: It’s the limit as compounding frequency → infinity

Compound Interest Time Calculations

How long to double your money?

Formula: \(2P = P(1 + r)^t\)
Simplify: \(2 = (1 + r)^t\)
Take logarithms: \(\ln(2) = t \cdot \ln(1 + r)\)
Solve: \(t = \frac{\ln(2)}{\ln(1 + r)}\)

. . .

Rule of 72: At r% interest, doubling time ≈ \(\frac{72}{r}\) years

Coffee Break - 15 Minutes

Advanced Algebraic Techniques

Solving Exponential Equations

Logarithms are the key tool

Solve \(3^{2x-1} = 81\)

Recognize: \(81 = 3^4\)
So: \(3^{2x-1} = 3^4\)
Therefore: \(2x - 1 = 4\)
Solve: \(x = 2.5\)

Solve \(5^x = 30\)

Take logarithms: \(\log(5^x) = \log(30)\)
Use power rule: \(x \cdot \log(5) = \log(30)\)
Solve: \(x = \frac{\log(30)}{\log(5)} \approx 2.113\)

Expanding Binomial Powers

Pascal’s Triangle

A pattern of binomial coefficients

Row 0:              1
Row 1:            1   1
Row 2:          1   2   1
Row 3:        1   3   3   1
Row 4:      1   4   6   4   1
Row 5:    1   5   10  10  5   1
Row 6:  1   6   15  20  15  6   1

Each number = sum of two above
Row n gives coefficients for \((a + b)^n\)
Symmetric pattern

Pair Exercise

Work together on binomial problems

Expand completely: \((x - 3)^3\)

Individual Exercise

Try to solve the following individually

If \(\log_2(x) + \log_4(x) = 3\), find x.
Expand: \((3x - 2y)^3\)
Simplify: \(\log_3(27) - \log_3(3)\)
Solve: \(2^{x+1} = 32\)

Wrap-up

Key Takeaways

Substitution is a powerful technique for simplifying expressions
Logarithms are inverse exponentials
The logarithm laws simplify complex calculations
Pascal’s triangle gives binomial coefficients
These tools are essential for calculus, statistics, and finance

For Next Time

Homework: Complete Tasks 01-05

Preview of Session 01-06 (Synthesis):

Integration of ALL Section 1 concepts
Complex problem-solving strategies
Business case studies

. . .

Start reviewing all Section 1 material - synthesis session next!

Questions & Discussion

Discussion

Your questions and insights are welcome!

Clarifications on logarithms?
Connections to other mathematical topics?
Applications you’re curious about?

See you next session!

The synthesis session will bring everything together!