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Lesson 2: Logarithmic Functions

Estimated time: 35-40 minutes

Learning Objectives

By the end of this lesson, you will be able to:

Understand logarithms as the inverse of exponential functions
Convert between exponential and logarithmic forms
Evaluate logarithms with various bases including common and natural logarithms
Apply the fundamental properties of logarithms
Graph logarithmic functions and identify key features
Use the change of base formula to evaluate logarithms with any base

Introduction: What is a Logarithm?

In the previous lesson, we learned about exponential functions like f(x) = 2^x. These functions answer the question: "What is 2 raised to the power of x?" But what if we want to reverse the question?

If 2^x = 8, what is x? We need a way to "undo" the exponential operation. This is where logarithms come in.

Logarithm: A logarithm is the inverse operation of exponentiation. It answers the question: "To what power must we raise the base to get a certain number?"

y = log_b(x) if and only if b^y = x

where b > 0, b ≠ 1, and x > 0

In the expression log_b(x):

b is called the base
x is called the argument
y (the result) is the exponent or logarithm

Reading Logarithmic Notation

log₂(8) = 3 is read as: "log base 2 of 8 equals 3"

This means: "2 raised to the 3rd power equals 8" or 2³ = 8

Important Restrictions:

The base b must be positive and not equal to 1 (b > 0, b ≠ 1)
The argument x must be positive (x > 0)
You cannot take the logarithm of zero or a negative number

Section 1: Evaluating Logarithms

The key to evaluating logarithms is to convert between exponential and logarithmic forms. Remember:

Exponential ↔ Logarithmic Conversion

b^y = x ⟺ log_b(x) = y

Example 1: Converting Exponential to Logarithmic Form

Convert each exponential equation to logarithmic form:

a) 2³ = 8

Solution:

Base: b = 2
Exponent: y = 3
Result: x = 8

Logarithmic form: log₂(8) = 3

b) 5² = 25

Solution:

Logarithmic form: log₅(25) = 2

c) 10^-2 = 0.01

Solution:

Logarithmic form: log₁₀(0.01) = -2

d) 3⁰ = 1

Solution:

Logarithmic form: log₃(1) = 0

Example 2: Converting Logarithmic to Exponential Form

Convert each logarithmic equation to exponential form:

a) log₄(16) = 2

Solution:

This means "4 to what power equals 16?" The answer is 2.

Exponential form: 4² = 16

b) log₅(125) = 3

Solution:

Exponential form: 5³ = 125

c) log₇(1) = 0

Solution:

Exponential form: 7⁰ = 1

d) log₂(1/8) = -3

Solution:

Exponential form: 2^-3 = 1/8

Example 3: Evaluating Logarithms

Evaluate each logarithm:

a) log₂(8)

Solution:

Ask: "2 to what power equals 8?"

Since 2³ = 8, we have log₂(8) = 3

b) log₃(27)

Solution:

Ask: "3 to what power equals 27?"

Since 3³ = 27, we have log₃(27) = 3

c) log₅(1/25)

Solution:

Ask: "5 to what power equals 1/25?"

Since 5^-2 = 1/25, we have log₅(1/25) = -2

d) log₁₀(1000)

Solution:

Ask: "10 to what power equals 1000?"

Since 10³ = 1000, we have log₁₀(1000) = 3

Special Logarithm Values

For any base b > 0, b ≠ 1:

log_b(1) = 0 because b⁰ = 1
log_b(b) = 1 because b¹ = b
log_b(bⁿ) = n for any real number n

Example 4: Using Special Logarithm Values

Evaluate each logarithm:

a) log₇(1)

Solution: Using the property log_b(1) = 0

log₇(1) = 0

b) log₉(9)

Solution: Using the property log_b(b) = 1

log₉(9) = 1

c) log₄(4⁵)

Solution: Using the property log_b(bⁿ) = n

log₄(4⁵) = 5

Example 5: Solving for x in Logarithmic Equations

Solve for x: log_x(16) = 2

Solution:

Step 1: Convert to exponential form

log_x(16) = 2 means x² = 16

Step 2: Solve for x

x² = 16

x = ±4

Step 3: Check restrictions

The base must be positive and not equal to 1, so x > 0

x = 4 (we reject x = -4)

Section 2: Common and Natural Logarithms

Two logarithm bases are so important that they have special notation:

Common Logarithm: Base 10 logarithm

log₁₀(x) = log(x)

When no base is written, the base is assumed to be 10.

Natural Logarithm: Base e logarithm (e ≈ 2.71828...)

log_e(x) = ln(x)

The natural logarithm uses the special notation "ln" instead of "log".

        Why These Logarithms Are Special
        Common logarithm (base 10): Used in pH calculations, decibels, Richter scale, and many scientific measurements
Natural logarithm (base e): Appears naturally in calculus, continuous growth/decay, and many mathematical formulas

      

Example 6: Evaluating Common Logarithms

a) log(1000)

Solution:

This means log₁₀(1000)

Ask: "10 to what power equals 1000?"

Since 10³ = 1000, we have log(1000) = 3

b) log(100)

Solution:

Since 10² = 100, we have log(100) = 2

c) log(0.01)

Solution:

Since 10^-2 = 0.01, we have log(0.01) = -2

d) log(1)

Solution:

Since 10⁰ = 1, we have log(1) = 0

Example 7: Evaluating Natural Logarithms

a) ln(e)

Solution:

This means log_e(e)

Using the property log_b(b) = 1

ln(e) = 1

b) ln(e³)

Solution:

Using the property log_b(bⁿ) = n

ln(e³) = 3

c) ln(1)

Solution:

Using the property log_b(1) = 0

ln(1) = 0

d) ln(1/e)

Solution:

Since 1/e = e^-1

ln(1/e) = ln(e^-1) = -1

Example 8: Using a Calculator for Logarithms

Most calculators cannot directly compute logarithms with arbitrary bases, but they can compute common (log) and natural (ln) logarithms.

a) log(50)

Solution:

Using a calculator: log(50) ≈ 1.699

b) ln(20)

Solution:

Using a calculator: ln(20) ≈ 2.996

c) log(0.5)

Solution:

Using a calculator: log(0.5) ≈ -0.301

Section 3: Properties of Logarithms

Logarithms have several important properties that make them powerful tools for solving problems. These properties follow directly from the laws of exponents.

Fundamental Logarithm Properties

For b > 0, b ≠ 1, and M, N > 0:

1. Product Rule: log_b(M · N) = log_b(M) + log_b(N)

2. Quotient Rule: log_b(M / N) = log_b(M) - log_b(N)

3. Power Rule: log_b(M^p) = p · log_b(M)

Why These Properties Work

Product Rule: If b^x = M and b^y = N, then M · N = b^x · b^y = b^x+y

Therefore, log_b(M · N) = x + y = log_b(M) + log_b(N)

Quotient Rule: If b^x = M and b^y = N, then M / N = b^x / b^y = b^x-y

Therefore, log_b(M / N) = x - y = log_b(M) - log_b(N)

Power Rule: If b^x = M, then M^p = (b^x)^p = b^xp

Therefore, log_b(M^p) = xp = p · log_b(M)

Example 9: Using the Product Rule to Expand

a) log₂(8x)

Solution:

log₂(8x) = log₂(8) + log₂(x)

Since log₂(8) = 3:

log₂(8x) = 3 + log₂(x)

b) ln(5xy)

Solution:

ln(5xy) = ln(5) + ln(x) + ln(y)

ln(5xy) = ln(5) + ln(x) + ln(y)

Example 10: Using the Quotient Rule to Expand

a) log(x/100)

Solution:

log(x/100) = log(x) - log(100)

Since log(100) = 2:

log(x/100) = log(x) - 2

b) ln(x/y)

Solution:

ln(x/y) = ln(x) - ln(y)

Example 11: Using the Power Rule

a) log₃(x⁵)

Solution:

log₃(x⁵) = 5 · log₃(x)

b) ln(√x)

Solution:

Since √x = x^1/2:

ln(√x) = ln(x^1/2) = (1/2) · ln(x)

ln(√x) = (1/2)ln(x)

c) log₂(1/x³)

Solution:

Since 1/x³ = x^-3:

log₂(1/x³) = log₂(x^-3) = -3 · log₂(x)

log₂(1/x³) = -3log₂(x)

Example 12: Expanding Complex Logarithmic Expressions

a) log₂(8x³)

Solution:

Step 1: Apply product rule

log₂(8x³) = log₂(8) + log₂(x³)

Step 2: Apply power rule to second term

= log₂(8) + 3log₂(x)

Step 3: Simplify log₂(8)

Since log₂(8) = 3:

log₂(8x³) = 3 + 3log₂(x)

b) ln(x²/y)

Solution:

Step 1: Apply quotient rule

ln(x²/y) = ln(x²) - ln(y)

Step 2: Apply power rule to first term

ln(x²/y) = 2ln(x) - ln(y)

Example 13: Condensing Logarithmic Expressions

Write as a single logarithm:

a) log(x) + 2log(y)

Solution:

Step 1: Apply power rule to second term

log(x) + 2log(y) = log(x) + log(y²)

Step 2: Apply product rule

log(x) + 2log(y) = log(xy²)

b) 3ln(x) - ln(y) + ln(z)

Solution:

Step 1: Apply power rule

3ln(x) - ln(y) + ln(z) = ln(x³) - ln(y) + ln(z)

Step 2: Apply product rule to first and third terms

= ln(x³z) - ln(y)

Step 3: Apply quotient rule

3ln(x) - ln(y) + ln(z) = ln(x³z/y)

Example 14: Using Properties to Evaluate

a) log₂(4) + log₂(8)

Solution:

Step 1: Apply product rule

log₂(4) + log₂(8) = log₂(4 · 8) = log₂(32)

Step 2: Evaluate

Since 2⁵ = 32:

log₂(4) + log₂(8) = 5

b) log₅(125) - log₅(5)

Solution:

Step 1: Apply quotient rule

log₅(125) - log₅(5) = log₅(125/5) = log₅(25)

Step 2: Evaluate

Since 5² = 25:

log₅(125) - log₅(5) = 2

Common Mistakes to Avoid:

log(M + N) ≠ log(M) + log(N) (There is no sum/difference property for addition/subtraction inside the log)
log(M - N) ≠ log(M) - log(N)
log(M/N) ≠ log(M) / log(N)
The properties only work when the operations are INSIDE the logarithm

Section 4: Graphs of Logarithmic Functions

Since logarithmic functions are inverses of exponential functions, their graphs are reflections of exponential graphs across the line y = x.

Basic Logarithmic Function: f(x) = log_b(x) where b > 1

Key Features:

Domain: (0, ∞) - only positive x-values
Range: (-∞, ∞) - all real numbers
x-intercept: (1, 0)
Vertical asymptote: x = 0 (the y-axis)
Passes through (b, 1)
Increasing if b > 1, decreasing if 0 < b < 1

Example 15: Graphing f(x) = log₂(x)

Solution:

Step 1: Identify key points

x	log₂(x)	Point
1/4	-2	(1/4, -2)
1/2	-1	(1/2, -1)
1	0	(1, 0)
2	1	(2, 1)
4	2	(4, 2)
8	3	(8, 3)

Step 2: Identify features

Vertical asymptote: x = 0
x-intercept: (1, 0)
Function increases slowly as x increases
Function decreases rapidly as x approaches 0 from the right

Example 16: Graphing f(x) = ln(x)

Solution:

Step 1: Identify key points

x	ln(x)	Point	Approximation
0.5	ln(0.5)	(0.5, -0.693)	≈ -0.693
1	ln(1)	(1, 0)	0
e ≈ 2.718	ln(e)	(e, 1)	1
3	ln(3)	(3, 1.099)	≈ 1.099
5	ln(5)	(5, 1.609)	≈ 1.609

Step 2: Features are similar to log₂(x)

Vertical asymptote: x = 0
x-intercept: (1, 0)
Passes through (e, 1)

Relationship Between Exponential and Logarithmic Graphs

The graphs of f(x) = b^x and g(x) = log_b(x) are reflections of each other across the line y = x.

If (a, b) is on the exponential graph, then (b, a) is on the logarithmic graph.

Example 17: Graphing f(x) = log(x) + 2 (Vertical Shift)

Solution:

Step 1: Start with the basic graph of log(x)

Step 2: Apply transformation

The "+2" shifts the entire graph UP by 2 units

Step 3: Identify new key points

Original point (1, 0) becomes (1, 2)
Original point (10, 1) becomes (10, 3)
Original point (100, 2) becomes (100, 4)

Step 4: Features

Vertical asymptote: still x = 0 (unchanged)
x-intercept: Solve log(x) + 2 = 0 → log(x) = -2 → x = 0.01
y-intercept: None (x = 0 not in domain)

Example 18: Graphing f(x) = log(x - 3) (Horizontal Shift)

Solution:

Step 1: Start with the basic graph of log(x)

Step 2: Apply transformation

The "-3" inside shifts the entire graph RIGHT by 3 units

Step 3: Identify new key points

Original point (1, 0) becomes (4, 0)
Original point (10, 1) becomes (13, 1)
Original point (100, 2) becomes (103, 2)

Step 4: Features

Vertical asymptote: x = 3 (shifted right by 3)
Domain: (3, ∞)
x-intercept: (4, 0)

Example 19: Determining the Equation from a Graph

A logarithmic function has a vertical asymptote at x = 0, passes through (5, 1), and has the general form f(x) = log_b(x). Find b.

Solution:

Step 1: Use the point (5, 1)

Since the function passes through (5, 1):

f(5) = 1

log_b(5) = 1

Step 2: Convert to exponential form

b¹ = 5

b = 5

Answer: f(x) = log₅(x)

Section 5: Change of Base Formula

Calculators typically only have buttons for log (base 10) and ln (base e). To evaluate logarithms with other bases, we use the change of base formula.

Change of Base Formula

For any positive numbers a, b, and x (where a ≠ 1 and b ≠ 1):

log_b(x) = log_a(x) / log_a(b)

Most commonly, we use a = 10 or a = e:

log_b(x) = log(x) / log(b) = ln(x) / ln(b)

Why the Change of Base Formula Works

Let y = log_b(x), which means b^y = x

Taking log_a of both sides:

log_a(b^y) = log_a(x)

Using the power rule:

y · log_a(b) = log_a(x)

Solving for y:

y = log_a(x) / log_a(b)

Example 20: Using Change of Base to Evaluate log₅(20)

Solution using base 10:

Step 1: Apply change of base formula

log₅(20) = log(20) / log(5)

Step 2: Use calculator

log(20) ≈ 1.301

log(5) ≈ 0.699

Step 3: Divide

log₅(20) ≈ 1.301 / 0.699 ≈ 1.861

Solution using base e:

log₅(20) = ln(20) / ln(5)

≈ 2.996 / 1.609

≈ 1.861 (same result)

Example 21: Evaluating log₂(50) Using a Calculator

Solution:

Step 1: Apply change of base formula

log₂(50) = log(50) / log(2)

Step 2: Use calculator

log(50) ≈ 1.699

log(2) ≈ 0.301

Step 3: Divide

log₂(50) ≈ 1.699 / 0.301 ≈ 5.644

Example 22: Comparing Logarithms with Different Bases

Which is larger: log₃(50) or log₇(50)?

Solution:

Step 1: Evaluate log₃(50)

log₃(50) = log(50) / log(3) ≈ 1.699 / 0.477 ≈ 3.561

Step 2: Evaluate log₇(50)

log₇(50) = log(50) / log(7) ≈ 1.699 / 0.845 ≈ 2.011

Step 3: Compare

log₃(50) > log₇(50)

Intuition: A smaller base requires a larger exponent to reach the same value.

Example 23: Solving for x Using Change of Base

Solve for x: log₃(x) = 2.5

Method 1: Convert to exponential form

log₃(x) = 2.5 means 3^2.5 = x

Using a calculator: 3^2.5 = 3^5/2 = √(3⁵) = √243

x ≈ 15.59

Method 2: Using change of base

log₃(x) = 2.5

log(x) / log(3) = 2.5

log(x) = 2.5 · log(3)

log(x) ≈ 2.5 · 0.477 ≈ 1.193

x = 10^1.193

x ≈ 15.59

Check Your Understanding

1. Convert to logarithmic form: 3⁴ = 81

Answer: log₃(81) = 4

Explanation: The base is 3, the exponent is 4, and the result is 81.

2. Convert to exponential form: log₆(216) = 3

Answer: 6³ = 216

Explanation: The logarithm tells us that 6 to the 3rd power equals 216.

3. Evaluate: log₄(64)

Answer: 3

Solution: Ask "4 to what power equals 64?" Since 4³ = 64, log₄(64) = 3

4. Evaluate: ln(e⁷)

Answer: 7

Explanation: Using the property log_b(bⁿ) = n, we have ln(e⁷) = 7

5. Expand using logarithm properties: log₅(25x²)

Solution:

log₅(25x²) = log₅(25) + log₅(x²)

= log₅(25) + 2log₅(x)

= 2 + 2log₅(x) (since log₅(25) = 2)

Answer: 2 + 2log₅(x)

6. Condense to a single logarithm: 4log(x) - 2log(y)

Solution:

4log(x) - 2log(y) = log(x⁴) - log(y²)

= log(x⁴/y²)

Answer: log(x⁴/y²)

7. What is the domain of f(x) = log(x - 5)?

Answer: (5, ∞)

Explanation: The argument of a logarithm must be positive, so x - 5 > 0, which means x > 5

8. What is the vertical asymptote of f(x) = ln(x + 3)?

Answer: x = -3

Explanation: The vertical asymptote occurs where the argument equals zero. x + 3 = 0 when x = -3

9. Use the change of base formula to evaluate log₄(30). Round to three decimal places.

Solution:

log₄(30) = log(30) / log(4)

≈ 1.477 / 0.602

Answer: ≈ 2.453

10. True or False: log(x + y) = log(x) + log(y)

Answer: FALSE

Explanation: This is a common mistake. The product rule states log(xy) = log(x) + log(y), not log(x + y). There is no property for the logarithm of a sum.

Key Takeaways

A logarithm is the inverse of an exponential function: y = log_b(x) if and only if b^y = x
The domain of a logarithmic function is (0, ∞) and the range is (-∞, ∞)
Special values: log_b(1) = 0, log_b(b) = 1, and log_b(bⁿ) = n
Common logarithm uses base 10: log(x) = log₁₀(x)
Natural logarithm uses base e: ln(x) = log_e(x)
Product rule: log_b(MN) = log_b(M) + log_b(N)
Quotient rule: log_b(M/N) = log_b(M) - log_b(N)
Power rule: log_b(M^p) = p · log_b(M)
Logarithmic graphs have a vertical asymptote at x = 0 and pass through (1, 0)
Change of base formula: log_b(x) = log(x) / log(b) = ln(x) / ln(b)

← Previous: Exponential Functions Next: Properties of Logarithms →

Lesson 2: Logarithmic Functions

Learning Objectives

Introduction: What is a Logarithm?

Reading Logarithmic Notation

Section 1: Evaluating Logarithms

Exponential ↔ Logarithmic Conversion

Example 1: Converting Exponential to Logarithmic Form

Example 2: Converting Logarithmic to Exponential Form

Example 3: Evaluating Logarithms

Special Logarithm Values

Example 4: Using Special Logarithm Values

Example 5: Solving for x in Logarithmic Equations

Section 2: Common and Natural Logarithms

Why These Logarithms Are Special

Example 6: Evaluating Common Logarithms

Example 7: Evaluating Natural Logarithms

Example 8: Using a Calculator for Logarithms

Section 3: Properties of Logarithms

Fundamental Logarithm Properties

Why These Properties Work

Example 9: Using the Product Rule to Expand

Example 10: Using the Quotient Rule to Expand

Example 11: Using the Power Rule

Example 12: Expanding Complex Logarithmic Expressions

Example 13: Condensing Logarithmic Expressions

Example 14: Using Properties to Evaluate

Section 4: Graphs of Logarithmic Functions

Example 15: Graphing f(x) = log2(x)

Example 16: Graphing f(x) = ln(x)

Relationship Between Exponential and Logarithmic Graphs

Example 17: Graphing f(x) = log(x) + 2 (Vertical Shift)

Example 18: Graphing f(x) = log(x - 3) (Horizontal Shift)

Example 19: Determining the Equation from a Graph

Section 5: Change of Base Formula

Change of Base Formula

Why the Change of Base Formula Works

Example 20: Using Change of Base to Evaluate log5(20)

Example 21: Evaluating log2(50) Using a Calculator

Example 22: Comparing Logarithms with Different Bases

Example 23: Solving for x Using Change of Base

Check Your Understanding

Key Takeaways

Example 15: Graphing f(x) = log₂(x)

Example 20: Using Change of Base to Evaluate log₅(20)

Example 21: Evaluating log₂(50) Using a Calculator