Logarithmic Functions

Understanding logarithms as inverse of exponentials, logarithmic properties, and solving logarithmic equations

Logarithmic Functions

What is a 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?"

Definition

log⁡b(x)=ymeansby=x\log_b(x) = y \quad \text{means} \quad b^y = x

Read as: "log base bb of xx equals yy"

Example: log⁡2(8)=3\log_2(8) = 3 because 23=82^3 = 8

Common Logarithms

  1. Common logarithm (base 10): log⁡(x)=log⁡10(x)\log(x) = \log_{10}(x)
  2. Natural logarithm (base ee): ln⁡(x)=log⁡e(x)\ln(x) = \log_e(x)

Converting Between Forms

| Exponential Form | Logarithmic Form | |------------------|------------------| | 23=82^3 = 8 | log⁡2(8)=3\log_2(8) = 3 | | 102=10010^2 = 100 | log⁡(100)=2\log(100) = 2 | | ex=5e^x = 5 | ln⁡(5)=x\ln(5) = x | | by=xb^y = x | log⁡b(x)=y\log_b(x) = y |

Properties of Logarithms

Product Rule

log⁡b(MN)=log⁡b(M)+log⁡b(N)\log_b(MN) = \log_b(M) + \log_b(N)

The log of a product is the sum of the logs.

Quotient Rule

log⁡b(MN)=log⁡b(M)−log⁡b(N)\log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N)

The log of a quotient is the difference of the logs.

Power Rule

log⁡b(Mp)=p⋅log⁡b(M)\log_b(M^p) = p \cdot \log_b(M)

The log of a power is the exponent times the log.

Change of Base Formula

log⁡b(x)=log⁡a(x)log⁡a(b)=ln⁡(x)ln⁡(b)\log_b(x) = \frac{\log_a(x)}{\log_a(b)} = \frac{\ln(x)}{\ln(b)}

Useful for calculating logs with different bases on a calculator.

Special Logarithm Values

  1. log⁡b(1)=0\log_b(1) = 0 because b0=1b^0 = 1
  2. log⁡b(b)=1\log_b(b) = 1 because b1=bb^1 = b
  3. log⁡b(bx)=x\log_b(b^x) = x (inverse property)
  4. blog⁡b(x)=xb^{\log_b(x)} = x (inverse property)

Logarithmic Functions

The function f(x)=log⁡b(x)f(x) = \log_b(x) has these properties:

  • Domain: (0,∞)(0, \infty) (only positive numbers)
  • Range: (−∞,∞)(-\infty, \infty) (all real numbers)
  • Vertical Asymptote: x=0x = 0 (the y-axis)
  • x-intercept: (1,0)(1, 0) since log⁡b(1)=0\log_b(1) = 0
  • Always increasing if b>1b > 1
  • Always decreasing if 0<b<10 < b < 1

Solving Logarithmic Equations

Strategy 1: Convert to exponential form

Strategy 2: Use logarithm properties to combine/simplify

Strategy 3: Check your answers (domain restrictions!)

Important Note

⚠️ You cannot take the log of a negative number or zero!

The domain of log⁡b(x)\log_b(x) requires x>0x > 0.

📚 Practice Problems

1Problem 1easy

❓ Question:

Evaluate the following logarithms: (a) log⁡3(27)\log_3(27), (b) log⁡5(125)\log_5(\frac{1}{25}), (c) log⁡(10000)\log(10000)

💡 Show Solution

Solution:

Part a) log⁡3(27)\log_3(27)

Ask: "3 raised to what power equals 27?"

3?=273^? = 27 33=273^3 = 27

Therefore: log⁡3(27)=3\log_3(27) = 3

Part b) log⁡5(125)\log_5\left(\frac{1}{25}\right)

Ask: "5 raised to what power equals 125\frac{1}{25}?"

5?=1255^? = \frac{1}{25}

We know 52=255^2 = 25, so 5−2=1255^{-2} = \frac{1}{25}

Therefore: log⁡5(125)=−2\log_5\left(\frac{1}{25}\right) = -2

Part c) log⁡(10000)\log(10000) (base 10)

Ask: "10 raised to what power equals 10000?"

10?=1000010^? = 10000 104=1000010^4 = 10000

Therefore: log⁡(10000)=4\log(10000) = 4

Answers: (a) 3, (b) -2, (c) 4

2Problem 2easy

❓ Question:

Evaluate the following logarithms without a calculator:

a) log⁡232\log_2 32 b) log⁡5125\log_5 \frac{1}{25} c) ln⁡e7\ln e^7

💡 Show Solution

Solution:

Part (a): log⁡232\log_2 32 means "2 to what power equals 32?"

25=322^5 = 32

Therefore: log⁡232=5\log_2 32 = 5

Part (b): log⁡5125\log_5 \frac{1}{25} means "5 to what power equals 125\frac{1}{25}?"

125=152=5−2\frac{1}{25} = \frac{1}{5^2} = 5^{-2}

Therefore: log⁡5125=−2\log_5 \frac{1}{25} = -2

Part (c): ln⁡e7\ln e^7 means log⁡ee7\log_e e^7

By the property log⁡bbx=x\log_b b^x = x:

ln⁡e7=7\ln e^7 = 7

3Problem 3medium

❓ Question:

Expand using logarithm properties: log⁡2(8x3y2)\log_2\left(\frac{8x^3}{y^2}\right)

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Solution:

Step 1: Apply the quotient rule.

log⁡2(8x3y2)=log⁡2(8x3)−log⁡2(y2)\log_2\left(\frac{8x^3}{y^2}\right) = \log_2(8x^3) - \log_2(y^2)

Step 2: Apply the product rule to the first term.

=log⁡2(8)+log⁡2(x3)−log⁡2(y2)= \log_2(8) + \log_2(x^3) - \log_2(y^2)

Step 3: Apply the power rule.

=log⁡2(8)+3log⁡2(x)−2log⁡2(y)= \log_2(8) + 3\log_2(x) - 2\log_2(y)

Step 4: Simplify log⁡2(8)\log_2(8).

Since 23=82^3 = 8, we have log⁡2(8)=3\log_2(8) = 3

=3+3log⁡2(x)−2log⁡2(y)= 3 + 3\log_2(x) - 2\log_2(y)

Answer: 3+3log⁡2(x)−2log⁡2(y)3 + 3\log_2(x) - 2\log_2(y)

4Problem 4medium

❓ Question:

Use properties of logarithms to expand the following expression completely:

log⁡3(x4yz2)\log_3 \left(\frac{x^4\sqrt{y}}{z^2}\right)

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Solution:

We'll use three key properties:

  • Product rule: log⁡b(MN)=log⁡bM+log⁡bN\log_b(MN) = \log_b M + \log_b N
  • Quotient rule: log⁡b(M/N)=log⁡bM−log⁡bN\log_b(M/N) = \log_b M - \log_b N
  • Power rule: log⁡bMp=plog⁡bM\log_b M^p = p\log_b M

Step 1: Apply quotient rule:

log⁡3(x4yz2)=log⁡3(x4y)−log⁡3(z2)\log_3 \left(\frac{x^4\sqrt{y}}{z^2}\right) = \log_3(x^4\sqrt{y}) - \log_3(z^2)

Step 2: Apply product rule to the first term:

=log⁡3(x4)+log⁡3(y)−log⁡3(z2)= \log_3(x^4) + \log_3(\sqrt{y}) - \log_3(z^2)

Step 3: Rewrite y=y1/2\sqrt{y} = y^{1/2} and apply power rule:

=4log⁡3x+12log⁡3y−2log⁡3z= 4\log_3 x + \frac{1}{2}\log_3 y - 2\log_3 z

Final answer: 4log⁡3x+12log⁡3y−2log⁡3z4\log_3 x + \frac{1}{2}\log_3 y - 2\log_3 z

5Problem 5easy

❓ Question:

Solve for xx: log⁡4(x)=3\log_4(x) = 3

💡 Show Solution

Solution:

Step 1: Convert from logarithmic form to exponential form.

log⁡4(x)=3⇒43=x\log_4(x) = 3 \quad \Rightarrow \quad 4^3 = x

Step 2: Evaluate.

x=43=64x = 4^3 = 64

Step 3: Check (optional but recommended).

log⁡4(64)=log⁡4(43)=3\log_4(64) = \log_4(4^3) = 3 ✓

Answer: x=64x = 64

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