The Logic of the Logarithm#
A logarithm answers one question: “What exponent do I need?”
- $2^? = 8$ → The answer is $\log_2 8 = 3$.
- $10^? = 1000$ → The answer is $\log_{10} 1000 = 3$.
- $5^? = 125$ → The answer is $\log_5 125 = 3$.
A logarithm is just an exponent. Every time you see $\log_b x$, read it as: “The power I must raise $b$ to in order to get $x$.”
1. The Conversion Between Forms#
This is the single most important thing to memorize:
$$ y = \log_b x \quad \Longleftrightarrow \quad b^y = x $$| Logarithmic Form | Exponential Form |
|---|---|
| $\log_2 8 = 3$ | $2^3 = 8$ |
| $\log_3 81 = 4$ | $3^4 = 81$ |
| $\log_{10} 0.01 = -2$ | $10^{-2} = 0.01$ |
The base stays the base. The log result is the exponent. The argument is the answer.
2. The General Form#
$$ y = a\log_b(x - p) + q $$3. The Parameters: What Each One Does#
The “$b$” Value — The Base#
| Value of $b$ | Effect |
|---|---|
| $b > 1$ | Graph is increasing (rises from left to right) |
| $0 < b < 1$ | Graph is decreasing (falls from left to right) |
| $b = 10$ | Common logarithm (written as $\log x$ on calculators) |
| $b = e \approx 2.718$ | Natural logarithm (written as $\ln x$) |
The “$a$” Value — Stretch and Reflection#
| Value of $a$ | Effect |
|---|---|
| $a > 0$ | Standard orientation |
| $a < 0$ | Graph is reflected in the horizontal — flipped upside down |
| $ | a |
| $0 < | a |
The “$p$” Value — Horizontal Shift (Vertical Asymptote)#
| Value of $p$ | Effect |
|---|---|
| $p > 0$ | Graph shifts right — asymptote moves to $x = p$ |
| $p < 0$ | Graph shifts left — asymptote moves to $x = p$ |
| $p = 0$ | Asymptote is the y-axis ($x = 0$) |
The vertical asymptote is always $x = p$.
This is because $\log_b(0)$ is undefined — you can never raise a positive base to any power and get zero.
The “$q$” Value — Vertical Shift#
| Value of $q$ | Effect |
|---|---|
| $q > 0$ | Graph shifts up |
| $q < 0$ | Graph shifts down |
4. Key Properties#
| Property | Value |
|---|---|
| Vertical Asymptote | $x = p$ |
| Domain | $x > p$ (if $a > 0$) or $x > p$ (argument must be positive) |
| Range | $y \in \mathbb{R}$ |
| x-intercept | Set $y = 0$: solve $0 = a\log_b(x - p) + q$ |
| y-intercept | Set $x = 0$: only exists if $p < 0$ (i.e., $0 > p$, so $0 - p > 0$) |
The “Anchor Point”#
For the basic log $y = \log_b x$, the graph always passes through $(1; 0)$ because $\log_b 1 = 0$ for any base.
For the shifted form $y = a\log_b(x - p) + q$, the anchor shifts to $(1 + p; q)$.
5. The Laws of Logarithms#
These laws are essential for simplifying and solving log equations. They all come from the laws of exponents.
Law 1: Product Rule#
$$ \log_b(m \times n) = \log_b m + \log_b n $$- Logic: When you multiply numbers, you add their exponents. Since logs ARE exponents, multiplying inside becomes adding outside.
- Example: $\log_2(4 \times 8) = \log_2 4 + \log_2 8 = 2 + 3 = 5$. Check: $\log_2 32 = 5$ ✓
Law 2: Quotient Rule#
$$ \log_b\left(\frac{m}{n}\right) = \log_b m - \log_b n $$- Logic: Division is the opposite of multiplication, so we subtract instead of add.
- Example: $\log_3\left(\frac{81}{3}\right) = \log_3 81 - \log_3 3 = 4 - 1 = 3$. Check: $\log_3 27 = 3$ ✓
Law 3: Power Rule#
$$ \log_b(m^k) = k \cdot \log_b m $$- Logic: An exponent on the argument can be “pulled down” as a multiplier. This is the most powerful law because it lets you solve for unknowns trapped inside exponents.
- Example: $\log_2(8^2) = 2 \cdot \log_2 8 = 2 \times 3 = 6$. Check: $\log_2 64 = 6$ ✓
Special Values#
| Expression | Value | Why |
|---|---|---|
| $\log_b 1$ | $0$ | Because $b^0 = 1$ |
| $\log_b b$ | $1$ | Because $b^1 = b$ |
| $\log_b b^k$ | $k$ | Because $b^k = b^k$ |
| $b^{\log_b x}$ | $x$ | The exponential and log “cancel out” |
6. Solving Logarithmic Equations#
Type 1: Direct Conversion#
Solve $\log_3 x = 4$
Convert to exponential form: $x = 3^4 = 81$
Type 2: Using Log Laws to Simplify#
Solve $\log_2 x + \log_2 (x - 2) = 3$
- Combine using the Product Rule: $\log_2[x(x-2)] = 3$
- Convert: $x(x-2) = 2^3 = 8$
- Expand: $x^2 - 2x - 8 = 0$
- Factor: $(x-4)(x+2) = 0$
- $x = 4$ or $x = -2$
- Check: $\log_2(-2)$ is undefined, so $x = -2$ is rejected.
- Answer: $x = 4$
Type 3: Exponential Equations Solved with Logs#
Solve $5^x = 20$
- Take $\log$ of both sides: $\log 5^x = \log 20$
- Apply the Power Rule: $x \log 5 = \log 20$
- Solve: $x = \frac{\log 20}{\log 5} = \frac{1.301}{0.699} \approx 1.86$
This technique is called the Change of Base method and is essential for solving exponential equations that can’t be reduced to the same base.
7. The Change of Base Formula#
$$ \log_b x = \frac{\log_a x}{\log_a b} $$Most commonly used with base 10 (your calculator’s log button):
This lets you evaluate any logarithm using your calculator, regardless of the base.
8. Relationship to the Exponential#
| Exponential $y = b^x$ | Logarithm $y = \log_b x$ |
|---|---|
| Passes through $(0; 1)$ | Passes through $(1; 0)$ |
| Asymptote: $y = 0$ (horizontal) | Asymptote: $x = 0$ (vertical) |
| Domain: $x \in \mathbb{R}$ | Domain: $x > 0$ |
| Range: $y > 0$ | Range: $y \in \mathbb{R}$ |
| Increasing if $b > 1$ | Increasing if $b > 1$ |
| Decreasing if $0 < b < 1$ | Decreasing if $0 < b < 1$ |
| Reflected across $y = x$ gives the log | Reflected across $y = x$ gives the exponential |
Worked Example: Full Exam-Style Question#
Given $f(x) = \log_3 x$
(a) Write down the domain and range of $f$.
- Domain: $x > 0$
- Range: $y \in \mathbb{R}$
(b) Write down the equation of $f^{-1}$.
Swap $x$ and $y$: $x = \log_3 y$ Convert to exponential: $y = 3^x$
$$ f^{-1}(x) = 3^x $$(c) Determine the value of $x$ for which $f(x) = -2$.
$$ \log_3 x = -2 $$$$ x = 3^{-2} = \frac{1}{9} $$(d) Sketch $f$ and $f^{-1}$ on the same set of axes.
Key points for $f$: $(1; 0)$, $(3; 1)$, $(9; 2)$, $(\frac{1}{3}; -1)$ Key points for $f^{-1}$: $(0; 1)$, $(1; 3)$, $(2; 9)$, $(-1; \frac{1}{3})$ Draw the line $y = x$ and verify they are reflections.
(e) For which values of $x$ is $f(x) \ge 1$?
$$ \log_3 x \ge 1 $$$$ x \ge 3^1 $$$$ x \ge 3 $$Worked Example: Using Log Laws#
Simplify without a calculator: $\log_2 48 - \log_2 3$
Using the Quotient Rule:
$$ = \log_2\left(\frac{48}{3}\right) = \log_2 16 = \log_2 2^4 = 4 $$Simplify: $2\log 5 + \log 4$
$$ = \log 5^2 + \log 4 = \log 25 + \log 4 = \log(25 \times 4) = \log 100 = 2 $$🚨 Common Mistakes#
- $\log(a + b) \ne \log a + \log b$: The Product Rule works for multiplication inside the log, NOT addition. There is no rule for $\log(a + b)$.
- Forgetting to reject negative arguments: After solving a log equation, always check that the arguments of ALL logs in the original equation are positive.
- $\log_b 0$ and $\log_b(\text{negative})$: Both are undefined. You cannot take the log of zero or a negative number.
- Confusing $\log$ and $\ln$: On your calculator,
log= base 10,ln= base $e$. In South African exams, $\log$ without a base typically means base 10. - The Power Rule direction: $\log_b m^k = k \log_b m$, NOT $(\log_b m)^k$. The exponent only comes down if it is on the argument, not on the entire log expression.
💡 Pro Tip: The “Definition” Shortcut#
If you ever get stuck on a log problem, convert it back to exponential form. $\log_b x = y$ means $b^y = x$. This simple conversion solves about 70% of log questions instantly.
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