Logarithms

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

Starting point: Years 9 and 10

What a logarithm is

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

$lo g_{a} y = x means a^{x} = y$

$lo g_{2} 8 = 3$ because $2^{3} = 8$ . Read it as "log base 2 of 8 is 3."

The log laws

These come straight from the index laws.

Solving with logs

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

$2^{x} = 10 \Rightarrow x = lo g_{2} 10 = \frac{lo g _{10} 10}{lo g _{10} 2} \approx 3.32$

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 11Harder log equations

In terms of other logs. If $lo g_{b} 2 = p$ and $lo g_{b} 7 = q$ , then $lo g_{b} 28 = lo g_{b} (2^{2} \times 7) = 2 p + q$ .

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

Different bases. Change every log to one base. $lo g_{4} x + lo g_{2} x = 6$ : since $lo g_{4} x = \frac{lo g _{2} x}{lo g _{2} 4} = \frac{1}{2} lo g_{2} x$ , we get $\frac{3}{2} lo g_{2} x = 6$ , so $lo g_{2} x = 4$ and $x = 16$ . Also $lo g_{x} 2 = \frac{1}{lo g _{2} x}$ .

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 $2^{x} - \frac{6}{2 ^{x}} = 1$ , 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 $x = lo g_{2} 3$ .

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

Chains collapse. By change of base, $lo g_{4} 3 \times lo g_{3} 2 = \frac{ln 3}{ln 4} \times \frac{ln 2}{ln 3} = \frac{ln 2}{ln 4} = \frac{1}{2}$ .

Making a letter the subject. From $2 lo g_{3} x - lo g_{3} y = 1$ : $lo g_{3} \frac{x ^{2}}{y} = 1$ , so $x^{2} = 3 y$ and $x = 3 y$ .

Quick summary

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

Quiz

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Try 3 questions now

What is $lo g_{2} 32$ ?
1. A 16
2. B 5
3. C 6
4. D 30
Show answer
B: 5
$2^{5} = 32$ .
What is $lo g_{10} 0.01$ ?
1. A 2
2. B $- 2$
3. C $- 1$
4. D 0.1
Show answer
B: $- 2$
$1 0^{- 2} = 0.01$ .
$lo g_{5} 1$ equals:
1. A 0
2. B 1
3. C 5
4. D undefined
Show answer
A: 0
$5^{0} = 1$ .