# Logarithms

> What a log is, the log laws, natural logs, log graphs and solving exponential equations.

From HSC Maths Advanced on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-advanced/logarithms/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The earthquake scale
News reports say a magnitude 6 earthquake hit somewhere overseas. A magnitude 7 a year later sounds only a little bigger. In fact it shakes the ground about **ten times** more.

The Richter scale uses **logarithms**. Each step up multiplies the shaking by 10. Sound in decibels and acidity in pH work the same way. Logarithms turn huge multiplying patterns into small, friendly numbers. They are also the tool that undoes an exponential.

**In short:** A logarithm is a power: $\log_a y = x$ means $a^x = y$. The log laws turn multiplying into adding, dividing into subtracting and powers into multiplying, and the natural log $\ln x = \log_e x$ undoes $e^x$.

## Starting point: Years 9 and 10

### What a logarithm is

A logarithm answers the question: **what power do I need?**

$$\log_a y = x \quad \text{means} \quad a^x = y$$

$\log_2 8 = 3$ because $2^3 = 8$. Read it as "log base 2 of 8 is 3."

### Worked example: Evaluating logs
- $\log_{10} 1000 = 3$ because $10^3 = 1000$.
- $\log_3 \dfrac{1}{9} = -2$ because $3^{-2} = \dfrac{1}{9}$.
- $\log_5 1 = 0$ because $5^0 = 1$.
- $\log_7 7 = 1$ because $7^1 = 7$.

### Calculator tip: The log keys
`log` is base 10 and `ln` is base $e$. For any other base, use the `log□□` key if your calculator has it, or the change of base rule: $\log_2 20 = \dfrac{\log 20}{\log 2} \approx 4.32$.

Check by going back: $2^{4.32} \approx 20$. Logs of zero or of a negative number give Math ERROR, because they do not exist.

### The log laws

These come straight from the index laws.

### Key idea: Log laws
$$\log_a(xy) = \log_a x + \log_a y$$
$$\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y$$
$$\log_a(x^n) = n\log_a x$$
$$\log_a 1 = 0 \qquad \log_a a = 1$$

### Worked example: Simplify
$\log_2 12 - \log_2 3 = \log_2 \dfrac{12}{3} = \log_2 4 = 2$.

$\log_{10} 25 + \log_{10} 4 = \log_{10} 100 = 2$.

### Solving with logs

When you cannot match bases, take logs of both sides.

$$2^x = 10 \;\Rightarrow\; x = \log_2 10 = \frac{\log_{10} 10}{\log_{10} 2} \approx 3.32$$

### Watch out: There is no law for the log of a sum
$\log(x + y)$ is **not** $\log x + \log y$. The addition law is for a **product**: $\log(xy)$.

### Exam tip: Change to index form
If a log question confuses you, rewrite it as a power. $\log_4 x = 3$ becomes $x = 4^3 = 64$.

**Mind map**

```text
Logarithms
  Meaning
    log_a y = x means aˣ = y
    What power do I need?
  Laws
    Product: add
    Quotient: subtract
    Power: bring down
  Natural log
    ln x = log_e x
    Undoes eˣ
  Graph
    Through (1, 0)
    Asymptote x = 0
  Uses
    Solve exponential equations
    d/dx ln x = 1/x
```

### Year 11: Natural logs and change of base
The **natural logarithm** uses base $e$:

$$\ln x = \log_e x$$

$\ln$ and $e$ undo each other: $\ln(e^x) = x$ and $e^{\ln x} = x$.

**Change of base.** To work out any log on a calculator:

$$\log_a x = \frac{\ln x}{\ln a} = \frac{\log_{10} x}{\log_{10} a}$$

**Solving.** Solve $5e^{2x} = 40$. Divide: $e^{2x} = 8$. Take ln: $2x = \ln 8$, so $x = \dfrac{\ln 8}{2} \approx 1.04$.

Solve $3^{x} = 2^{x+1}$. Take ln: $x\ln 3 = (x + 1)\ln 2$, so $x(\ln 3 - \ln 2) = \ln 2$ and $x = \dfrac{\ln 2}{\ln 1.5} \approx 1.71$.

### Year 11: Harder log equations
**In terms of other logs.** If $\log_b 2 = p$ and $\log_b 7 = q$, then $\log_b 28 = \log_b(2^2 \times 7) = 2p + q$.

**Check the answers.** Logs need positive inputs, so test every answer in the original equation. Solve $\log_2 x + \log_2(x + 2) = 3$: $\log_2\big(x(x + 2)\big) = 3$, so $x^2 + 2x = 8$ and $x = 2$ or $x = -4$. Only $x = 2$ works, because $\log_2(-4)$ does not exist. Similarly $2\log x = \log(x + 6)$ becomes $x^2 = x + 6$, giving $x = 3$ (reject $-2$).

**Different bases.** Change every log to one base. $\log_4 x + \log_2 x = 6$: since $\log_4 x = \dfrac{\log_2 x}{\log_2 4} = \dfrac{1}{2}\log_2 x$, we get $\dfrac{3}{2}\log_2 x = 6$, so $\log_2 x = 4$ and $x = 16$. Also $\log_x 2 = \dfrac{1}{\log_2 x}$.

**Quadratics in disguise.** In $(\ln x)^2 - \ln x - 2 = 0$, let $u = \ln x$: $(u - 2)(u + 1) = 0$, so $x = e^2$ or $x = e^{-1}$. In $2^x - \dfrac{6}{2^x} = 1$, multiply by $2^x$ and let $u = 2^x$: $u^2 - u - 6 = 0$ gives $u = 3$ (reject $u = -2$, because $2^x > 0$), so $x = \log_2 3$.

**Two log equations.** $\log(xy) = 3$ and $\log\dfrac{x}{y} = 1$ give $\log x + \log y = 3$ and $\log x - \log y = 1$. Add: $\log x = 2$, so $x = 100$ and $y = 10$.

**Chains collapse.** By change of base, $\log_4 3 \times \log_3 2 = \dfrac{\ln 3}{\ln 4} \times \dfrac{\ln 2}{\ln 3} = \dfrac{\ln 2}{\ln 4} = \dfrac{1}{2}$.

**Making a letter the subject.** From $2\log_3 x - \log_3 y = 1$: $\log_3\dfrac{x^2}{y} = 1$, so $x^2 = 3y$ and $x = \sqrt{3y}$.

### Year 11: Graphs of logarithms
$y = \log_a x$ is the reflection of $y = a^x$ in the line $y = x$, because logs and exponentials undo each other.

- Passes through $(1, 0)$.
- Vertical asymptote $x = 0$.
- Domain $x > 0$. You cannot take the log of zero or a negative.
- Grows very slowly.

$y = \ln(x - 2)$ is moved right 2, with asymptote $x = 2$.

### Year 12: Calculus with ln
$$\frac{d}{dx}\ln x = \frac{1}{x} \qquad \frac{d}{dx}\ln f(x) = \frac{f'(x)}{f(x)}$$

So $\dfrac{d}{dx}\ln(x^2 + 1) = \dfrac{2x}{x^2 + 1}$.

Going backwards, $\displaystyle\int \frac{1}{x}\,dx = \ln|x| + C$.

**Half-life.** A drug decays as $M = 50e^{-0.2t}$. When is half left? $25 = 50e^{-0.2t}$, so $e^{-0.2t} = 0.5$, $-0.2t = \ln 0.5$, and $t = \dfrac{\ln 2}{0.2} \approx 3.47$ hours.

## Quick summary

- $\log_a y = x$ means $a^x = y$. A log is a power.
- Product: add logs. Quotient: subtract logs. Power: bring the power to the front.
- $\log_a 1 = 0$ and $\log_a a = 1$.
- $\ln x = \log_e x$ undoes $e^x$. Change of base: $\log_a x = \dfrac{\ln x}{\ln a}$.
- $y = \log_a x$ passes through $(1, 0)$ with asymptote $x = 0$.

## Quiz

Try these questions from the chapter quiz.

1. What is $\log_2 32$?
   - A. 16
   - B. 5
   - C. 6
   - D. 30
   - Answer: B. 5. $2^5 = 32$.

2. What is $\log_{10} 0.01$?
   - A. 2
   - B. $-2$
   - C. $-1$
   - D. 0.1
   - Answer: B. $-2$. $10^{-2} = 0.01$.

3. $\log_5 1$ equals:
   - A. 0
   - B. 1
   - C. 5
   - D. undefined
   - Answer: A. 0. $5^0 = 1$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-advanced/logarithms/#quiz