6.2 Logarithm

Detailed Theory: Logarithms

1. Basic Concepts and Definitions

1.1 What is a Logarithm?

A logarithm is the inverse operation to exponentiation.

If $b^x = y$, then $\log_b y = x$

Reading: "log base b of y equals x"

Example: Since $2^3 = 8$, then $\log_2 8 = 3$

1.2 Components of a Logarithm

In $\log_b a = c$:

• b is the base (must be positive and not equal to 1)

• a is the argument (must be positive)

• c is the value or exponent

1.3 Why Use Logarithms?

1. Simplify calculations: Convert multiplication to addition

2. Solve exponential equations

3. Model exponential growth/decay

4. Scale large ranges (Richter scale, pH, decibels)

1.4 Exponential-Logarithmic Relationship

Logarithm and exponentiation are inverse operations:

$$b^{\log_b x} = x \quad \text{and} \quad \log_b(b^x) = x$$

Important: These only work when $x > 0$

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2. Types of Logarithms

2.1 Common Logarithms (Base 10)

• Base = 10

• Notation: $\log x$ (base 10 implied)

• Also called Briggsian logarithms

• Used in scientific calculations

Example: $\log 100 = 2$ because $10^2 = 100$

2.2 Natural Logarithms (Base e)

• Base = $e \approx 2.71828$

• Notation: $\ln x$

• Also called Napierian logarithms

• Used in calculus and higher mathematics

Example: $\ln e = 1$ because $e^1 = e$

2.3 Binary Logarithms (Base 2)

• Base = 2

• Notation: $\log_2 x$ or $\text{lb } x$

• Used in computer science and information theory

Example: $\log_2 8 = 3$ because $2^3 = 8$

2.4 Relationship Between Different Bases

For any positive numbers $a, b, x$ (with $a, b \neq 1$):

$$\log_a x = \frac{\log_b x}{\log_b a}$$

This is called the change of base formula.

Special case: $\log_a x = \frac{\ln x}{\ln a} = \frac{\log x}{\log a}$

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3. Fundamental Properties and Laws

3.1 Basic Properties

a) Logarithm of 1

For any base $b > 0$, $b \neq 1$:

$$\log_b 1 = 0$$

Reason: $b^0 = 1$

b) Logarithm of Base

$$\log_b b = 1$$

Reason: $b^1 = b$

c) Logarithm of Power of Base

$$\log_b (b^k) = k$$

Example: $\log_2 (2^5) = 5$

d) Power of Logarithm

$$b^{\log_b x} = x$$

Example: $10^{\log 100} = 100$

3.2 Three Fundamental Laws

a) Product Law

$$\log_b (mn) = \log_b m + \log_b n$$

Verification: Let $\log_b m = x$ and $\log_b n = y$

Then $m = b^x$ and $n = b^y$

So $mn = b^x \cdot b^y = b^{x+y}$

Thus $\log_b (mn) = x + y = \log_b m + \log_b n$

Example: $\log_2 (4 \times 8) = \log_2 4 + \log_2 8 = 2 + 3 = 5$

b) Quotient Law

$$\log_b \left(\frac{m}{n}\right) = \log_b m - \log_b n$$

Verification: Let $\log_b m = x$ and $\log_b n = y$

Then $m = b^x$ and $n = b^y$

So $\frac{m}{n} = \frac{b^x}{b^y} = b^{x-y}$

Thus $\log_b \left(\frac{m}{n}\right) = x - y = \log_b m - \log_b n$

Example: $\log_3 \left(\frac{27}{9}\right) = \log_3 27 - \log_3 9 = 3 - 2 = 1$

c) Power Law

$$\log_b (m^n) = n \log_b m$$

Verification: Let $\log_b m = x$

Then $m = b^x$

So $m^n = (b^x)^n = b^{nx}$

Thus $\log_b (m^n) = nx = n \log_b m$

Example: $\log_2 (8^3) = 3 \log_2 8 = 3 \times 3 = 9$

3.3 Additional Important Properties

a) Change of Base

$$\log_a b = \frac{1}{\log_b a}$$

Proof: Using change of base formula:

$$\log_a b = \frac{\log_b b}{\log_b a} = \frac{1}{\log_b a}$$

b) Chain Rule

$$\log_a b \times \log_b c = \log_a c$$

Proof:

$$\log_a b \times \log_b c = \frac{\log b}{\log a} \times \frac{\log c}{\log b} = \frac{\log c}{\log a} = \log_a c$$

c) Root Property

$$\log_b \sqrt[n]{m} = \frac{1}{n} \log_b m$$

Reason: $\sqrt[n]{m} = m^{1/n}$, so apply power law

Example: $\log \sqrt{100} = \frac{1}{2} \log 100 = \frac{1}{2} \times 2 = 1$

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4. Domain and Range

4.1 Domain Restrictions

For $\log_b x$ to be defined:

1. Base b: $b > 0$ and $b \neq 1$

   • If $b = 1$: $1^x$ is always 1, so inverse not unique

   • If $b = 0$: $0^x$ is 0 for $x > 0$, undefined for $x \leq 0$

   • If $b < 0$: Results become complex

2. Argument x: $x > 0$

   • You cannot take log of zero or negative numbers in real numbers

4.2 Range

For $\log_b x$:

• If $b > 1$: Range is all real numbers ($-\infty, \infty$)

• If $0 < b < 1$: Range is all real numbers ($-\infty, \infty$)

4.3 Example: Find Domain of $\log_2 (x-3)$

Argument must be positive: $x - 3 > 0$

So $x > 3$

Domain: $(3, \infty)$

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5. Graphs of Logarithmic Functions

5.1 Basic Shape

The graph of $y = \log_b x$ has:

1. Vertical asymptote at $x = 0$ (y-axis)

2. x-intercept at $x = 1$ (since $\log_b 1 = 0$)

3. Passes through point $(b, 1)$ (since $\log_b b = 1$)

5.2 Two Cases

Case 1: Base b > 1

• Function is increasing

• As $x \to 0^+$, $y \to -\infty$

• As $x \to \infty$, $y \to \infty$

• Example: $y = \log_2 x$, $y = \ln x$

Case 2: Base 0 < b < 1

• Function is decreasing

• As $x \to 0^+$, $y \to \infty$

• As $x \to \infty$, $y \to -\infty$

• Example: $y = \log_{0.5} x$

5.3 Transformations

a) Vertical Shift

$y = \log_b x + c$ shifts graph up by c units

b) Horizontal Shift

$y = \log_b (x - h)$ shifts graph right by h units

c) Vertical Stretch/Compression

$y = a \log_b x$ multiplies y-values by a

d) Horizontal Stretch/Compression

$y = \log_b (kx)$ affects x-values

5.4 Example: Graph $y = 2\log_3 (x-1) + 1$

1. Start with basic $y = \log_3 x$

2. Shift right 1 unit: $y = \log_3 (x-1)$

3. Vertical stretch by 2: $y = 2\log_3 (x-1)$

4. Shift up 1 unit: $y = 2\log_3 (x-1) + 1$

Domain: $x-1 > 0 \Rightarrow x > 1$

Vertical asymptote: $x = 1$

x-intercept: Set $y=0$: $2\log_3 (x-1) + 1 = 0$

$\log_3 (x-1) = -\frac{1}{2}$

$x-1 = 3^{-1/2} = \frac{1}{\sqrt{3}}$

$x = 1 + \frac{1}{\sqrt{3}}$

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6. Solving Logarithmic Equations

6.1 Basic Methods

a) Using Definition

Convert to exponential form.

Example: Solve $\log_2 x = 3$

By definition: $x = 2^3 = 8$

b) Using Properties

Combine logs using product/quotient/power laws.

Example: Solve $\log x + \log (x-3) = 1$

Using product law: $\log [x(x-3)] = 1$

So $x(x-3) = 10^1 = 10$

$x^2 - 3x - 10 = 0$

$(x-5)(x+2) = 0$

$x = 5$ or $x = -2$

But domain requires $x > 0$ and $x-3 > 0$, so $x > 3$

Thus $x = 5$ is only solution.

c) Using Change of Base

Make all logs have same base.

Example: Solve $\log_2 x = \log_4 9$

Change base: $\log_4 9 = \frac{\log_2 9}{\log_2 4} = \frac{\log_2 9}{2}$

So equation becomes: $\log_2 x = \frac{\log_2 9}{2}$

Multiply by 2: $2\log_2 x = \log_2 9$

$\log_2 (x^2) = \log_2 9$

Thus $x^2 = 9$, so $x = 3$ or $x = -3$

Since $x > 0$, $x = 3$

6.2 Checking for Extraneous Solutions

IMPORTANT: Always check solutions in original equation!

Reasons for extraneous solutions:

1. Argument of log becomes non-positive

2. Base becomes invalid

Example: Solve $\log_2 (x-1) + \log_2 (x+1) = 3$

Using product law: $\log_2 [(x-1)(x+1)] = 3$

$(x-1)(x+1) = 2^3 = 8$

$x^2 - 1 = 8$

$x^2 = 9$

$x = 3$ or $x = -3$

Check domain: $x-1 > 0$ and $x+1 > 0$

For $x = 3$: $3-1=2>0$, $3+1=4>0$ ✓

For $x = -3$: $-3-1=-4<0$ ✗

So only solution is $x = 3$

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7. Solving Exponential Equations Using Logarithms

7.1 Basic Method

To solve $a^x = b$:

1. Take log of both sides: $\log(a^x) = \log b$

2. Use power law: $x \log a = \log b$

3. Solve for x: $x = \frac{\log b}{\log a}$

7.2 Examples

Example 1: Solve $2^x = 5$

Take log: $\log(2^x) = \log 5$

$x \log 2 = \log 5$

$x = \frac{\log 5}{\log 2} \approx \frac{0.6990}{0.3010} \approx 2.322$

Example 2: Solve $3^{2x-1} = 7^{x+2}$

Take natural log: $\ln(3^{2x-1}) = \ln(7^{x+2})$

$(2x-1)\ln 3 = (x+2)\ln 7$

Expand: $2x\ln 3 - \ln 3 = x\ln 7 + 2\ln 7$

Group x terms: $2x\ln 3 - x\ln 7 = 2\ln 7 + \ln 3$

Factor: $x(2\ln 3 - \ln 7) = 2\ln 7 + \ln 3$

$x = \frac{2\ln 7 + \ln 3}{2\ln 3 - \ln 7}$

Example 3: Solve $e^{2x} - 5e^x + 6 = 0$

Let $y = e^x$, then equation becomes:

$y^2 - 5y + 6 = 0$

$(y-2)(y-3) = 0$

So $y = 2$ or $y = 3$

Thus $e^x = 2$ or $e^x = 3$

$x = \ln 2$ or $x = \ln 3$

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8. Logarithmic Inequalities

8.1 Important Consideration

When solving $\log_b f(x) > \log_b g(x)$:

Case 1: If b > 1 (function increasing) Then $f(x) > g(x)$ AND $f(x) > 0$, $g(x) > 0$

Case 2: If 0 < b < 1 (function decreasing) Then $f(x) < g(x)$ AND $f(x) > 0$, $g(x) > 0$

8.2 Examples

Example 1: Solve $\log_2 (x-1) < 3$

Since base 2 > 1, inequality direction preserved:

First, domain: $x-1 > 0 \Rightarrow x > 1$

Inequality: $\log_2 (x-1) < 3$

Convert: $x-1 < 2^3 = 8$

So $x < 9$

Combining with domain: $1 < x < 9$

Example 2: Solve $\log_{0.5} (2x-1) > -1$

Base 0.5 is between 0 and 1, so inequality reverses:

First, domain: $2x-1 > 0 \Rightarrow x > \frac{1}{2}$

Inequality: $\log_{0.5} (2x-1) > -1$

Since base < 1, reverse: $2x-1 < (0.5)^{-1} = 2$

So $2x-1 < 2 \Rightarrow 2x < 3 \Rightarrow x < \frac{3}{2}$

Combining with domain: $\frac{1}{2} < x < \frac{3}{2}$

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9. Applications of Logarithms

9.1 Scientific Applications

a) Richter Scale (Earthquakes)

Magnitude $M = \log_{10} \left(\frac{A}{A_0}\right)$

where A is amplitude, $A_0$ is reference amplitude.

Example: Earthquake 1000 times stronger than reference:

$M = \log_{10} 1000 = 3$

b) pH Scale (Acidity)

$\text{pH} = -\log_{10} [\text{H}^+]$

where $[\text{H}^+]$ is hydrogen ion concentration.

Example: Solution with $[\text{H}^+] = 10^{-5}$ M:

$\text{pH} = -\log_{10} (10^{-5}) = -(-5) = 5$

c) Decibel Scale (Sound)

$\text{dB} = 10 \log_{10} \left(\frac{I}{I_0}\right)$

where I is intensity, $I_0$ is reference intensity.

Example: Sound 100 times more intense than reference:

$\text{dB} = 10 \log_{10} 100 = 10 \times 2 = 20 \text{ dB}$

9.2 Mathematical Applications

a) Solving Compound Interest

For principal P, rate r, time t, compounded n times per year:

$A = P\left(1 + \frac{r}{n}\right)^{nt}$

To find time to reach amount A:

$t = \frac{\log(A/P)}{n \log(1 + r/n)}$

b) Half-life Problems

For exponential decay: $N = N_0 e^{-kt}$

Half-life $t_{1/2} = \frac{\ln 2}{k}$

c) Logarithmic Scales

• Used when data spans many orders of magnitude

• Examples: star magnitudes, financial charts (log scale)

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10. Natural Logarithms and Calculus

10.1 Definition of e

The number e is defined as:

$$e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n \approx 2.71828$$

10.2 Derivative of ln x

$$\frac{d}{dx} (\ln x) = \frac{1}{x}, \quad x > 0$$

10.3 Integral of 1/x

$$\int \frac{1}{x} \, dx = \ln |x| + C$$

10.4 Derivatives of General Logarithms

Using change of base: $\log_b x = \frac{\ln x}{\ln b}$

So $\frac{d}{dx} (\log_b x) = \frac{1}{x \ln b}$

10.5 Logarithmic Differentiation

Technique for differentiating complicated functions:

Steps:

1. Take ln of both sides: $\ln y = \ln f(x)$

2. Differentiate implicitly: $\frac{y'}{y} = \frac{d}{dx}[\ln f(x)]$

3. Solve for y'

Example: Find derivative of $y = x^x$

Take ln: $\ln y = \ln(x^x) = x \ln x$

Differentiate: $\frac{y'}{y} = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$

So $y' = y(\ln x + 1) = x^x (\ln x + 1)$

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11. Logarithmic Series and Limits

11.1 Series Expansion for ln(1+x)

For $-1 < x \leq 1$:

$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n}$$

11.2 Important Limits

1. $\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$

2. $\lim_{x \to \infty} \frac{\ln x}{x} = 0$

3. $\lim_{x \to 0^+} x \ln x = 0$

4. $\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e$

11.3 Example: Evaluate $\lim_{x \to 0} \frac{\ln(\cos x)}{x^2}$

Using series: $\cos x = 1 - \frac{x^2}{2} + \cdots$

So $\ln(\cos x) = \ln\left(1 - \frac{x^2}{2} + \cdots\right) \approx -\frac{x^2}{2}$ for small x

Thus $\lim_{x \to 0} \frac{\ln(\cos x)}{x^2} = \lim_{x \to 0} \frac{-x^2/2}{x^2} = -\frac{1}{2}$

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12. Common Mistakes and Pitfalls

12.1 Most Common Errors

1. $\log_b (m+n) \neq \log_b m + \log_b n$

   • Correct: $\log_b (mn) = \log_b m + \log_b n$

2. $\log_b (m-n) \neq \log_b m - \log_b n$

   • Correct: $\log_b \left(\frac{m}{n}\right) = \log_b m - \log_b n$

3. $(\log_b m)^n \neq \log_b (m^n)$

   • Correct: $\log_b (m^n) = n \log_b m$

4. $\log_b m \times \log_b n \neq \log_b (mn)$

   • Correct: $\log_b m + \log_b n = \log_b (mn)$

5. Forgetting domain restrictions

   • Argument must be positive

   • Base must be positive and not 1

12.2 Example of Common Error

Wrong: $\log_2 (8 + 4) = \log_2 8 + \log_2 4 = 3 + 2 = 5$

Right: $\log_2 (8 + 4) = \log_2 12 = \log_2 (4 \times 3) = \log_2 4 + \log_2 3 = 2 + \log_2 3 \approx 3.585$

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13. Solved Examples for Practice

Example 1: Simplify $\frac{\log_2 27}{\log_2 3}$

Solution:

Using change of base: $\log_2 27 = \log_2 (3^3) = 3\log_2 3$

So $\frac{\log_2 27}{\log_2 3} = \frac{3\log_2 3}{\log_2 3} = 3$

Example 2: Solve $2\log x = \log 4 + \log 16$

Solution:

Right side: $\log 4 + \log 16 = \log (4 \times 16) = \log 64$

So $2\log x = \log 64$

$\log (x^2) = \log 64$

Thus $x^2 = 64$

$x = 8$ or $x = -8$

Since argument must be positive: $x = 8$

Example 3: Evaluate $\log_3 54 - \log_3 8 + \log_3 4$

Solution:

Using properties: $\log_3 54 - \log_3 8 + \log_3 4 = \log_3 \left(\frac{54 \times 4}{8}\right)$

$= \log_3 \left(\frac{216}{8}\right) = \log_3 27 = \log_3 (3^3) = 3$

Example 4: If $\log_{10} 2 = 0.3010$, find $\log_{10} 8$

Solution:

$\log_{10} 8 = \log_{10} (2^3) = 3\log_{10} 2 = 3 \times 0.3010 = 0.9030$

Example 5: Solve $2^{x+1} = 5^{x-1}$

Solution:

Take log: $\log(2^{x+1}) = \log(5^{x-1})$

$(x+1)\log 2 = (x-1)\log 5$

Expand: $x\log 2 + \log 2 = x\log 5 - \log 5$

Group x terms: $x\log 2 - x\log 5 = -\log 5 - \log 2$

$x(\log 2 - \log 5) = -(\log 5 + \log 2)$

$x = \frac{-(\log 5 + \log 2)}{\log 2 - \log 5} = \frac{\log 5 + \log 2}{\log 5 - \log 2} = \frac{\log 10}{\log(5/2)}$

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14. Important Formulas Summary

14.1 Fundamental Laws

1. Product: $\log_b (mn) = \log_b m + \log_b n$

2. Quotient: $\log_b \left(\frac{m}{n}\right) = \log_b m - \log_b n$

3. Power: $\log_b (m^n) = n \log_b m$

14.2 Special Values

1. $\log_b 1 = 0$

2. $\log_b b = 1$

3. $\log_b (b^k) = k$

4. $b^{\log_b x} = x$

14.3 Change of Base

$$\log_a x = \frac{\log_b x}{\log_b a} = \frac{\ln x}{\ln a} = \frac{\log x}{\log a}$$

14.4 Domain and Range

• Domain: $x > 0$

• Base: $b > 0$, $b \neq 1$

• Range: All real numbers

14.5 Common Logarithms

• $\log 10 = 1$

• $\log 100 = 2$

• $\log 1000 = 3$

• $\log 0.1 = -1$

• $\log 0.01 = -2$

14.6 Natural Logarithms

• $\ln e = 1$

• $\ln 1 = 0$

• $\ln e^x = x$

• $e^{\ln x} = x$

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15. Exam Tips and Strategies

15.1 Problem-Solving Approach

1. Identify the type: Equation, inequality, simplification, application?

2. Check domain: Ensure arguments are positive

3. Use properties: Combine/expand using laws

4. Convert forms: Exponential ↔ Logarithmic

5. Verify solution: Check in original equation

15.2 Common Exam Questions

1. Simplify expressions using logarithmic properties

2. Solve equations involving logs and exponents

3. Prove identities using logarithmic laws

4. Application problems (pH, Richter scale, compound interest)

5. Graph transformations of logarithmic functions

15.3 Quick Checks

1. For $\log_b x$: Is $x > 0$? Is $b > 0$ and $b \neq 1$?

2. When solving: Did you consider all possible solutions?

3. For inequalities: Did you check if base > 1 or 0 < base < 1?

4. For applications: Are units consistent?

15.4 Memory Aids

• Product law: "Log of product = Sum of logs"

• Quotient law: "Log of quotient = Difference of logs"

• Power law: "Exponent comes down as multiplier"

• Domain: "Can't take log of zero or negative"

This comprehensive theory covers all aspects of logarithms with detailed explanations and examples, providing complete preparation for the entrance examination.