Logarithms
Complete Notes & Formulae for Twelfth Grade (Precalculus)
1. Convert Between Exponential and Logarithmic Form
Definition:
A logarithm is the inverse of an exponential function. It answers the question: "To what power must we raise the base to get this number?"
\[ b^x = y \quad \Leftrightarrow \quad \log_b y = x \]
Read as: "log base b of y equals x"
Key Components:
• b = base (must be positive, b ≠ 1)
• x = exponent (logarithm value)
• y = result/argument (must be positive)
Conversion Examples:
| Exponential Form | Logarithmic Form |
|---|---|
| \( 2^3 = 8 \) | \( \log_2 8 = 3 \) |
| \( 10^2 = 100 \) | \( \log_{10} 100 = 2 \) |
| \( 5^0 = 1 \) | \( \log_5 1 = 0 \) |
| \( 3^{-2} = \frac{1}{9} \) | \( \log_3 \frac{1}{9} = -2 \) |
| \( e^x = y \) | \( \ln y = x \) |
Common Logarithms:
• Common log: \( \log x = \log_{10} x \) (base 10)
• Natural log: \( \ln x = \log_e x \) (base e ≈ 2.718)
2. Evaluate Logarithms
Basic Evaluation:
Method:
Ask yourself: "What power of the base gives the argument?"
\[ \log_b y = x \text{ means } b^x = y \]
Examples:
Evaluate: \( \log_2 16 \)
Ask: 2 to what power equals 16?
\( 2^4 = 16 \)
Therefore: \( \log_2 16 = 4 \)
Evaluate: \( \log_5 \frac{1}{25} \)
Ask: 5 to what power equals 1/25?
\( 5^{-2} = \frac{1}{25} \)
Therefore: \( \log_5 \frac{1}{25} = -2 \)
Special Values:
• \( \log_b 1 = 0 \) (any base to the 0 power = 1)
• \( \log_b b = 1 \) (base to the 1st power = itself)
• \( \log_b b^x = x \) (inverse property)
3. Change of Base Formula
Formula:
\[ \log_b a = \frac{\log_c a}{\log_c b} \]
where c is any positive base (commonly 10 or e)
Common Forms:
\[ \log_b a = \frac{\log a}{\log b} \quad \text{(using base 10)} \]
\[ \log_b a = \frac{\ln a}{\ln b} \quad \text{(using base e)} \]
When to Use:
• To evaluate logarithms with bases other than 10 or e on a calculator
• To compare logarithms with different bases
• To simplify expressions with multiple bases
Example:
Evaluate: \( \log_5 20 \)
Using change of base formula:
\( \log_5 20 = \frac{\log 20}{\log 5} = \frac{1.301}{0.699} \approx 1.861 \)
Or using natural log:
\( \log_5 20 = \frac{\ln 20}{\ln 5} \approx 1.861 \)
4. Product Property of Logarithms
Property:
The logarithm of a product equals the sum of the logarithms
\[ \log_b(MN) = \log_b M + \log_b N \]
Examples:
• \( \log_2(8 \times 4) = \log_2 8 + \log_2 4 = 3 + 2 = 5 \)
• \( \ln(xy) = \ln x + \ln y \)
• \( \log_3(9 \times 27) = \log_3 9 + \log_3 27 = 2 + 3 = 5 \)
5. Quotient Property of Logarithms
Property:
The logarithm of a quotient equals the difference of the logarithms
\[ \log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N \]
Examples:
• \( \log_2\left(\frac{32}{4}\right) = \log_2 32 - \log_2 4 = 5 - 2 = 3 \)
• \( \ln\left(\frac{x}{y}\right) = \ln x - \ln y \)
• \( \log\left(\frac{100}{10}\right) = \log 100 - \log 10 = 2 - 1 = 1 \)
6. Power Property of Logarithms
Property:
The logarithm of a power equals the exponent times the logarithm
\[ \log_b(M^p) = p \cdot \log_b M \]
Examples:
• \( \log_2(8^3) = 3 \log_2 8 = 3(3) = 9 \)
• \( \ln(x^5) = 5 \ln x \)
• \( \log(100^2) = 2 \log 100 = 2(2) = 4 \)
• \( \log_3(\sqrt{9}) = \log_3(9^{1/2}) = \frac{1}{2}\log_3 9 = \frac{1}{2}(2) = 1 \)
7. Properties of Logarithms: Mixed Review
All Properties Summary:
| Property | Formula | Description |
|---|---|---|
| Product | \( \log_b(MN) = \log_b M + \log_b N \) | Log of product = sum of logs |
| Quotient | \( \log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N \) | Log of quotient = difference of logs |
| Power | \( \log_b(M^p) = p \log_b M \) | Log of power = exponent × log |
| Change of Base | \( \log_b a = \frac{\log_c a}{\log_c b} \) | Convert to different base |
| Identity | \( \log_b b^x = x \) | Inverse property |
| Zero Property | \( \log_b 1 = 0 \) | Log of 1 is always 0 |
| Base Property | \( \log_b b = 1 \) | Log base of itself is 1 |
Combined Example:
Expand: \( \log_2\left(\frac{x^3y}{z^2}\right) \)
Step 1: Apply quotient property:
\( \log_2(x^3y) - \log_2(z^2) \)
Step 2: Apply product property to first term:
\( \log_2(x^3) + \log_2 y - \log_2(z^2) \)
Step 3: Apply power property:
\( 3\log_2 x + \log_2 y - 2\log_2 z \)
8. Evaluate Logarithms Using Properties
Strategy:
1. Break down complex arguments using properties
2. Simplify to logarithms you can evaluate
3. Use known values to find the answer
Example 1 (Condensing):
Condense: \( 2\log_3 x + \log_3 y - 3\log_3 z \)
Step 1: Apply power property:
\( \log_3(x^2) + \log_3 y - \log_3(z^3) \)
Step 2: Apply product property:
\( \log_3(x^2y) - \log_3(z^3) \)
Step 3: Apply quotient property:
\( \log_3\left(\frac{x^2y}{z^3}\right) \)
Example 2 (Evaluating):
Given: \( \log_2 3 = 1.585 \) and \( \log_2 5 = 2.322 \)
Find: \( \log_2 15 \)
Recognize: \( 15 = 3 \times 5 \)
Apply product property:
\( \log_2 15 = \log_2(3 \times 5) = \log_2 3 + \log_2 5 \)
\( = 1.585 + 2.322 = 3.907 \)
Example 3 (Complex):
Evaluate: \( \log_2 32 - \log_2 4 + \log_2 8 \)
Method 1: Evaluate each separately
\( = 5 - 2 + 3 = 6 \)
Method 2: Combine first
\( = \log_2\left(\frac{32 \times 8}{4}\right) = \log_2 64 = 6 \)
9. Quick Reference Summary
Essential Formulas:
Definition: \( b^x = y \Leftrightarrow \log_b y = x \)
Product: \( \log_b(MN) = \log_b M + \log_b N \)
Quotient: \( \log_b\left(\frac{M}{N}\right) = \log_b M - \log_b N \)
Power: \( \log_b(M^p) = p\log_b M \)
Change of Base: \( \log_b a = \frac{\log a}{\log b} \)
Special: \( \log_b 1 = 0 \), \( \log_b b = 1 \), \( \log_b b^x = x \)
📚 Study Tips
✓ Logarithms are the inverse of exponential functions
✓ Product property turns multiplication into addition
✓ Quotient property turns division into subtraction
✓ Power property brings exponents down as coefficients
✓ Use change of base formula for calculator evaluations