Exponentials and logarithms: meaning and identities

Powers first

For a positive base $a$ and any real exponents $m$ and $n$:

Rule	Example
$a^m \cdot a^n = a^{m+n}$	$2^3 \cdot 2^4 = 2^7$
$\dfrac{a^m}{a^n} = a^{m-n}$	$\dfrac{5^6}{5^4} = 25$
$(a^m)^n = a^{mn}$	$(3^2)^5 = 3^{10}$
$(ab)^n = a^n b^n$	$6^3 = 2^3 \cdot 3^3$
$a^0 = 1$, $a^{-n} = \dfrac{1}{a^n}$	$4^{-2} = \frac{1}{16}$
$a^{\frac{m}{n}} = \sqrt[n]{a^m}$	$8^{\frac{2}{3}} = \sqrt[3]{64} = 4$

There is no rule for $a^m + a^n$; the best you can do is factor, as in $2^{x+1} + 2^x = 2^x(2 + 1) = 3 \cdot 2^x$. A negative exponent makes a reciprocal, never a negative number. With a fractional exponent, take the root first when you can: $27^{\frac{4}{3}} = (\sqrt[3]{27})^4 = 3^4 = 81$, which keeps the numbers small.

What a logarithm is

$\log_a b$ is the exponent you must put on $a$ to get $b$:

$$\log_a b = c \iff a^c = b \qquad (a > 0,\ a \neq 1,\ b > 0).$$

So $\log_2 32 = 5$ because $2^5 = 32$, $\log_{10} 0.001 = -3$ because $10^{-3} = 0.001$, and $\log_9 3 = \frac{1}{2}$ because $9^{\frac{1}{2}} = 3$. The base must be positive and different from 1, and the argument must be positive: $\log_2(-4)$ and $\log_2 0$ do not exist. The natural logarithm $\ln$ has base $e \approx 2.718$.

Two identities follow straight from the definition, because a logarithm and a power with the same base undo each other:

$$a^{\log_a x} = x \quad (x > 0), \qquad \log_a(a^x) = x.$$

Also $\log_a 1 = 0$ and $\log_a a = 1$ for every valid base.

The logarithm rules

Because logarithms are exponents, the power rules turn into these, for positive $x$ and $y$:

Rule	Comes from
$\log_a(xy) = \log_a x + \log_a y$	multiplying powers adds exponents
$\log_a \dfrac{x}{y} = \log_a x - \log_a y$	dividing powers subtracts exponents
$\log_a(x^k) = k \log_a x$	a power of a power multiplies exponents
$\log_a b = \dfrac{\log_c b}{\log_c a}$	change of base, for any valid base $c$

The change-of-base rule is how you evaluate logarithms whose base and argument are powers of the same number: $\log_8 4 = \frac{\log_2 4}{\log_2 8} = \frac{2}{3}$. It also gives $\log_a b = \frac{1}{\log_b a}$.

What the rules do not say matters as much. $\log_a(x + y)$ has no simplification. $\frac{\log_a x}{\log_a y}$ is not $\log_a \frac{x}{y}$. $(\log_a x)^2$ is not $2 \log_a x$. Wrong options are built from exactly these.

Estimating and comparing without a calculator

A logarithm can be placed between two integers by bracketing its argument with powers of the base: $2^3 = 8 < 10 < 16 = 2^4$, so $3 < \log_2 10 < 4$.

To compare large powers, rewrite them with a common exponent or a common base. For example $2^{30} = (2^3)^{10} = 8^{10}$ and $3^{20} = (3^2)^{10} = 9^{10}$, so $3^{20} > 2^{30}$ even though $30 > 20$. When neither works, split off a shared factor, or compare logarithms instead of the numbers, since $\log$ with base greater than 1 preserves order.

Expressing one logarithm through others

A common question gives $\log_a 2$ and $\log_a 3$ as letters and asks for $\log_a 72$. Factor the argument into the given numbers and apply the rules: $72 = 2^3 \cdot 3^2$, so $\log_a 72 = 3\log_a 2 + 2\log_a 3$. Fractions and decimals work the same way, with subtraction for the denominator.