Maths1_Week6_Logarithms

Maths 1 Week 6: Logarithmic Functions

0. Prerequisites

NOTE

What you need to know:

• Exponents: $b^y=x$.
• Roots: $\sqrt[y]{x}$.
• Basic Algebra: Solving for $x$.

Quick Refresher

• Logarithm: The inverse of an exponent.
  • ${\log }_{b}x=y⟺b^y=x$.
  • Question: “To what power must I raise $b$ to get $x$?”
• Common Log: $\log x$ (Base 10).
• Natural Log: $\ln x$ (Base $e\approx 2.718$).

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1. Core Concepts

1.1 Properties of Logarithms

1. Product Rule: ${\log }_b(xy)={\log }_{b}x+{\log }_{b}y$.
   • Mnemonic: Multiplication inside becomes Addition outside.
2. Quotient Rule: ${\log }_b(x/y)={\log }_{b}x-{\log }_{b}y$.
   • Mnemonic: Division inside becomes Subtraction outside.
3. Power Rule: ${\log }_b(x^k)=k{\log }_{b}x$.
   • Mnemonic: Bring the power down to the front.
4. Identity: ${\log }_{b}b=1$, ${\log }_{b}1=0$.

1.2 Graph & Domain

• Domain: You can ONLY take the log of a positive number.
  • Inside $>0$.
  • Vertical Asymptote at $x=0$ (or where argument is 0).
• Range: All real numbers ($\mathbb{R}$).

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2. Pattern Analysis & Goated Solutions

Pattern 1: Solving Exponential Equations

Context: “Solve $5^x=17$.”

TIP

Mental Algorithm:

1. Isolate: Get the exponential part alone ($b^x=\dots$).
2. Log Both Sides: Take $\ln$ or $\log$ of both sides.
3. Power Rule: Bring $x$ down ($x\ln b=\ln \dots$).
4. Solve: Divide to find $x$.

Example (Detailed Solution)

Problem: Solve $3^{x+1}=7$. Solution:

1. Log Both Sides:
   • $\ln (3^{x+1})=\ln (7)$.
2. Power Rule:
   • $(x+1)\ln 3=\ln 7$.
3. Isolate x:
   • $x+1=\frac{\ln 7}{\ln 3}$.
   • $x=\frac{\ln 7}{\ln 3}-1$.
   • Note: $\frac{\ln 7}{\ln 3}={\log }_{3}7$. Answer: $x={\log }_{3}7-1$.

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Pattern 2: Solving Logarithmic Equations

Context: “Solve ${\log }_{2}x+{\log }_2(x-2)=3$.”

TIP

Mental Algorithm:

1. Combine: Use Product/Quotient rules to get a single log on one side. ${\log }_b(\dots )=A$.
2. Exponentiate: Rewrite as $b^A=(\dots )$.
3. Solve: Solve the resulting algebraic equation.
4. Check: CRITICAL STEP. Plug answers back into original equation. Argument must be $>0$.

Example (Detailed Solution)

Problem: Solve ${\log }_{2}x+{\log }_2(x-2)=3$. Solution:

1. Combine:
   • ${\log }_2(x\cdot (x-2))=3$.
   • ${\log }_2(x^2-2x)=3$.
2. Exponentiate:
   • $2^3=x^2-2x$.
   • $8=x^2-2x$.
3. Solve:
   • $x^2-2x-8=0$.
   • Factors of -8 adding to -2: (-4, 2).
   • $(x-4)(x+2)=0$.
   • $x=4$ or $x=-2$.
4. Check:
   • Try $x=4$: ${\log }_{2}4$ (OK), ${\log }_{2}2$ (OK). Valid.
   • Try $x=-2$: ${\log }_2(-2)$ (ERROR). Invalid. Answer: $x=4$.

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Pattern 3: Domain of Log Functions

Context: “Find domain of $f(x)=\log (x^2-9)$.”

TIP

Mental Algorithm:

1. Set Condition: Argument $>0$.
2. Solve Inequality: Use Wavy Curve method if quadratic.

Example (Detailed Solution)

Problem: Find domain of $y=\ln (x^2-5x+6)$. Solution:

1. Condition: $x^2-5x+6>0$.
2. Factor: $(x-2)(x-3)>0$.
3. Roots: 2, 3.
4. Test Regions:
   • $x>3$ (e.g., 4): $(+)(+)=+$ (Good).
   • $2<x<3$ (e.g., 2.5): $(+)(-)=-$ (Bad).
   • $x<2$ (e.g., 1): $(-)(-)=+$ (Good). Answer: $(-\infty ,2)\cup (3,\infty )$.

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3. Practice Exercises

1. Expand: Expand $\ln (\frac{x^2}{y})$.
   • Hint: $2\ln x-\ln y$.
2. Solve: $e^{2x}=5$.
   • Hint: $2x=\ln 5⟹x=(\ln 5)/2$.
3. Domain: $\log (x+5)$.
   • Hint: $x+5>0⟹x>-5$.

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🧠 Level Up: Advanced Practice

Question 1: Hidden Domain Constraints

Problem: Solve ${\log }_2(x-4)+{\log }_2(x-2)=3$. Logic:

1. Combine: ${\log }_2((x-4)(x-2))=3$.
2. Exponentiate: $(x-4)(x-2)=2^3=8$.
3. Solve Quadratic: $x^2-6x+8=8⟹x^2-6x=0⟹x(x-6)=0$.
4. Potential Roots: $x=0,x=6$.
5. Check Domain:
   • For $\log (x-4)$, we need $x>4$.
   • $x=0$ fails ($0-4=-4$).
   • $x=6$ works ($6-4=2,6-2=4$). Answer: $x=6$.

Question 2: Inequality Flip Trap

Problem: Solve ${\log }_{0.5}(x^2-3)>{\log }_{0.5}(1)$. Logic:

1. Base Check: Base is $0.5$ (between 0 and 1).
2. Flip Inequality: $x^2-3<1$.
3. Solve: $x^2<4⟹-2<x<2$.
4. Domain Constraint: Argument must be positive. $x^2-3>0⟹x^2>3$.
   • $x>\sqrt{3}$ or $x<-\sqrt{3}$.
5. Intersection:
   • $(-2,2)$ AND $(-\infty ,-\sqrt{3})\cup (\sqrt{3},\infty )$.
   • Approx $\sqrt{3}\approx 1.732$.
   • Intervals: $(-2,-\sqrt{3})\cup (\sqrt{3},2)$. Answer: $(-2,-\sqrt{3})\cup (\sqrt{3},2)$.

Question 3: Change of Base

Problem: Solve ${\log }_2(x)+{\log }_4(x)+{\log }_{16}(x)=7$. Logic:

1. Convert to Base 2:
   • ${\log }_4(x)=\frac{{\log }_{2}x}{{\log }_{2}4}=\frac{1}{2}{\log }_{2}x$.
   • ${\log }_{16}(x)=\frac{{\log }_{2}x}{{\log }_{2}16}=\frac{1}{4}{\log }_{2}x$.
2. Equation: ${\log }_{2}x(1+1/2+1/4)=7$.
3. Sum Fractions: $1+0.5+0.25=1.75=7/4$.
4. Solve: $\frac{7}{4}{\log }_{2}x=7⟹{\log }_{2}x=4$.
5. Result: $x=2^4=16$. Answer: 16.