The laws come from the index laws

x = a^m,\quad y = a^n

xy = a^{m+n} \;\Rightarrow\; \log_a xy = \log_a x + \log_a y

\tfrac{x}{y} = a^{m-n} \;\Rightarrow\; \log_a \tfrac{x}{y} = \log_a x - \log_a y

x^k = a^{km} \;\Rightarrow\; \log_a x^k = k\log_a x

every log law is an index law, read backwards

Logarithms

A logarithm is a power, asked backwards.

$2^4$ asks: what is 2 to the power 4? A logarithm asks the question the other way round — what power of 2 gives 16? On the graph of $y = 2^x$ , that means reading from the height back to the input.

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The definition

\log_a b = x \iff a^x = b

$2^4 = 16$ , so $\log_2 16 = 4$ . Likewise $\log_{10} 1000 = 3$ , and $\log_5 \tfrac{1}{25} = -2$ — fractions below 1 have negative logs.

The product law

\log_a xy = \log_a x + \log_a y

On a scale where every step doubles, multiplying is adding lengths: $4 \times 8$ lands on 32 because $2 + 3 = 5$ . It's just $2^2 \times 2^3 = 2^5$ , seen through the log.

Quotient and power laws

Dividing is subtracting lengths: $\log_2 40 - \log_2 5 = \log_2 8 = 3$ . And a power comes out in front: $\log_a x^k = k\log_a x$ . Every log law is an index law, read backwards.

Solving for a power

a^x = b \;\Rightarrow\; x = \frac{\log b}{\log a}

When the unknown is the power, take logs of both sides: the power law brings it down where you can reach it. $3^x = 20$ gives $x = 2.73$ — exactly where $y = 3^x$ crosses $y = 20$ .

Chapter 10

Logs turn powers into something you can hold.

Combining logs, exponents that are expressions, equations with logs in them, and twelve graded exam questions with full worked solutions are waiting.

Read the full chapter › Jump to the questions ›