Logarithm Rules

A log is an exponent, so the log laws are the exponent laws backwards: _b(ac) = _b a + _b c, and friends.

The idea

Definition (Logarithm).

Let $b > 0$ with $b \ne 1$, and let $a > 0$. The logarithm $\log_b a$ is the exponent to which $b$ must be raised to give $a$: $\log_b a = x$ means $b^{x} = a$.

So $\log_2 8 = 3$, because $2^{3} = 8$. Two values follow at once: $\log_b 1 = 0$, since $b^{0} = 1$, and $\log_b b = 1$, since $b^{1} = b$.

Theorem (Logarithm rules).

For any base $b > 0$ with $b \ne 1$, any positive numbers $a$ and $c$, and any real number $n$, $\log_b(ac) = \log_b a + \log_b c,$ $\log_b\!\left(\frac{a}{c}\right) = \log_b a - \log_b c,$ $\log_b(a^{n}) = n\log_b a.$

The three rules are the exponent laws read backwards. Write $a = b^{x}$ and $c = b^{y}$, so that $x = \log_b a$ and $y = \log_b c$. Multiplying the numbers adds their exponents, $b^{x}b^{y} = b^{x+y}$, which is the product rule. Dividing subtracts them, $b^{x}/b^{y} = b^{x-y}$, which is the quotient rule. Raising to a power multiplies them, $(b^{x})^{n} = b^{nx}$, which is the power rule.

Each rule steps an operation down one level — a product becomes a sum, a quotient a difference, a power a product. This is what logarithms are for: they trade hard arithmetic for easier arithmetic, and they bring an unknown down from an exponent so we can solve for it as an ordinary unknown.

Ways to work on it

• Walkthrough. Derive the product, quotient, and power rules from the exponent laws, then apply each to clean powers of 2 and 3.
• Practice. Evaluate a logarithm with the product rule, or combine two same-base logs into one and read off the value.
• Hardest. Simplify a log expression to a single integer, or solve an equation built from logarithms.

Not sure where to start? Take the ten-question placement test.