10. Properties of logarithms — Real numbers

Conditions

Throughout this section:

a, b \in (0,1) \cup (1,+\infty) \;\text{(bases)} \qquad c, d \in (0,+\infty) \qquad n \in \mathbb{N}

In fact, property (3) below works for any real exponent, not just natural numbers: in the homework we use it with $n = \frac{1}{2}$ or $n = -2$.

Basic properties

Property	Why
$\log_{a} 0$ does not exist	no power $a^{c}$ equals $0$
$\log_{a} 1 = 0$	$a^{0} = 1$
$\log_{a} a = 1$	$a^{1} = a$
$\log_{a} a^{n} = n$	the power you raise $a$ to in order to get $a^{n}$ is $n$
$\log_{a} (-c)$ does not exist	$a^{x}$ is always positive

Operational properties

The three we will use

\begin{aligned} &\text{(1)} && \log_{a} c + \log_{a} d = \log_{a} (c \cdot d) \\[4pt] &\text{(2)} && \log_{a} c - \log_{a} d = \log_{a} \frac{c}{d} \\[4pt] &\text{(3)} && \log_{a} c^{n} = n \cdot \log_{a} c \end{aligned}

The logarithm turns products into sums, quotients into differences and powers into products.

Where (1) comes from

If $\log_{a} c = x$ and $\log_{a} d = y$, then $a^{x} = c$ and $a^{y} = d$. Multiplying:

c \cdot d = a^{x} \cdot a^{y} = a^{x+y} \;\Longrightarrow\; \log_{a}(c \cdot d) = x + y = \log_{a} c + \log_{a} d

It is the power rule $a^{x} \cdot a^{y} = a^{x+y}$ read backwards. (2) and (3) follow in the same way from $a^{x} : a^{y} = a^{x-y}$ and $(a^{x})^{n} = a^{nx}$.

Careful: there is no property for a sum inside. $\log_{a}(c + d) \neq \log_{a} c + \log_{a} d$. For example, $\log(10 + 10) = \log 20 \approx 1.301$, but $\log 10 + \log 10 = 2$.

Change of base

Change-of-base formula

\log_{a} b = \frac{\log b}{\log a} = \frac{\ln b}{\ln a}

It lets you work out a logarithm in any base on the calculator, using the log or ln keys.

Examples

Example 1

\log_{3} 9 + \log_{3} 243 \overset{(1)}{=} \log_{3}(9 \cdot 243) = \log_{3}\left(3^{2} \cdot 3^{5}\right) = \log_{3} 3^{7} \overset{(3)}{=} 7 \cdot \log_{3} 3 = 7 \cdot 1 = 7

Example 2

2\log x - 3\log x \overset{(3)}{=} \log x^{2} - \log x^{3} \overset{(2)}{=} \log \frac{x^{2}}{x^{3}} = \log \frac{1}{x} = \log x^{-1} \overset{(3)}{=} -\log x

Example 3: an exponential equation

When the unknown is in the exponent, take logarithms of both sides and property (3) brings it down:

3^{x} = 5 \;\Longrightarrow\; \log_{3} 3^{x} = \log_{3} 5 \;\overset{(3)}{\Longrightarrow}\; x \cdot \log_{3} 3 = \log_{3} 5 \;\Longrightarrow\; x = \log_{3} 5

And the change of base lets us compute it:

x = \log_{3} 5 = \frac{\ln 5}{\ln 3} \approx \frac{1.609}{1.099} \approx 1.465

Check: $3^{1.465} \approx 5.00$.