Logarithms: definition and laws

A log­a­rithm answers the ques­tion "to what power must I raise this base to get that num­ber?" Log­a­rithms turn mul­ti­pli­ca­tion into addi­tion and pow­ers into mul­ti­pli­ca­tion, which is why they mat­ter so much. In this chap­ter they appear as a new kind of func­tion, $f(x) = \log_a x$, and to work with that func­tion you need the def­i­n­i­tion, the rea­sons behind the con­di­tions on the base, and the stan­dard laws. This les­son col­lects all three.

Log­a­rithms from pow­ers#

Look at these pow­ers of $2$ and read each one back­wards.

The log­a­rithm is the expo­nent. "$\log_2 4 = 2$" says exactly the same thing as "$2^2 = 4$".

The def­i­n­i­tion#

$$a^x = N \iff x = \log_a N$$

where $a > 0$, $a \neq 1$, $N > 0$ and $x \in \mathbb{R}$. Here $a$ is called the base.

So the base lies in $(0, 1) \cup (1, \infty) = (0, \infty) - \{1\}$, the num­ber $N$ lies in $(0, \infty)$, and the log­a­rithm $x$ can be any real num­ber in $(-\infty, \infty)$.

Why the base has these con­di­tions#

Each for­bid­den base leads to non­sense.

The base can­not be $0$. We have $0^2 = 0$ and also $0^3 = 0$. If $\log_0 0$ existed, it would have to equal both $2$ and $3$ at once, which is impos­si­ble. So $a \neq 0$.

The base can­not be $1$. We have $1^2 = 1$ and $1^3 = 1$. Then $\log_1 1$ would have to be $2$ and $3$ at the same time. So $a \neq 1$.

The base can­not be neg­a­tive. We have $(-1)^2 = 1$ and $(-1)^4 = 1$, so $\log_{-1} 1$ would have to be both $2$ and $4$. Worse, $(-1)^{1/2}$ is not even a real num­ber. So $a$ is never neg­a­tive.

That leaves $a \in (0, 1) \cup (1, \infty)$. And since a pos­i­tive base raised to any real power is pos­i­tive, $N = a^x$ is always pos­i­tive: log­a­rithms of $0$ and of neg­a­tive num­bers do not exist.

Exam­ple 1#

Change between the two forms.

(a) $2^4 = 16$ in log form is $\log_2 16 = 4$.

(b) If $\log_2 x = 5$, then $x = 2^5 = 32$.

(c) If $\log_5 (x - 1) = -1$, then $\displaystyle x - 1 = 5^{-1} = \frac{1}{5}$, so $\displaystyle x = \frac{6}{5}$.

Exam­ple 2#

Eval­u­ate (a) $\log_2 32$ (b) $\log_{1/5} 25$ (c) $\log_{1/3} 3$ (d) $\log_3 9$.

(a) $2^5 = 32$, so $\log_2 32 = 5$.

(b) $\displaystyle \left(\frac{1}{5}\right)^{-2} = (5^{-1})^{-2} = 5^2 = 25$, so $\log_{1/5} 25 = -2$.

(c) $\displaystyle \left(\frac{1}{3}\right)^{-1} = 3$, so $\log_{1/3} 3 = -1$.

(d) $3^2 = 9$, so $\log_3 9 = 2$.

The laws of log­a­rithms#

In each law below, every base is pos­i­tive and not $1$, and every num­ber whose log­a­rithm is taken is pos­i­tive.

1. $\log_a 1 = 0$, because $a^0 = 1$. For exam­ple, $\log_3 1 = 0$. But $\log_1 1$ and $\log_{-1} 1$ are unde­fined, because $1$ and $-1$ are not allowed as bases.

2. $\log_a a = 1$, because $a^1 = a$. For exam­ple, $\log_2 2 = 1$ and $\displaystyle \log_{1/5} \frac{1}{5} = 1$.

3. Prod­uct law. $\log_a (mn) = \log_a m + \log_a n$.

Proof. Let $\log_a m = x$ and $\log_a n = y$. Then $m = a^x$ and $n = a^y$, so

$$mn = a^x \cdot a^y = a^{x + y}$$

Read­ing this in log form, $\log_a (mn) = x + y = \log_a m + \log_a n$.

4. Quo­tient law. $\displaystyle \log_a \frac{m}{n} = \log_a m - \log_a n$. Com­bin­ing laws 3 and 4,

$$\displaystyle \log_a \frac{mn}{pq} = \log_a m + \log_a n - \log_a p - \log_a q$$

5. Power law. $\displaystyle \log_{b^n} a^m = \frac{m}{n} \log_b a$. Two spe­cial cases are used most:

$$\displaystyle \log_b a^m = m \log_b a, \qquad \log_{b^n} a = \frac{1}{n} \log_b a$$

6. $\displaystyle \log_b a = \frac{1}{\log_a b}$.

7. $\log_b a \cdot \log_a b = 1$.

8. Chain law. $\log_b a \cdot \log_c b = \log_c a$. It extends to longer chains:

$$\log_b a \cdot \log_c b \cdot \log_d c = \log_d a$$

9. Change of base. $\displaystyle \log_b a = \frac{\log a}{\log b} = \frac{\log_x a}{\log_x b}$, for any allowed base $x$.

Do not con­fuse this with the quo­tient law: $\displaystyle \frac{\log a}{\log b}$ is not equal to $\displaystyle \log \frac{a}{b}$.

10. $a^{\log_a N} = N$. For exam­ple, $2^{\log_2 3} = 3$. This is just the def­i­n­i­tion read in one line: $\log_a N$ is the power that turns $a$ into $N$.

11. $a^{\log_c b} = b^{\log_c a}$. For exam­ple, $2^{\log_3 5} = 5^{\log_3 2}$. The base and the num­ber can swap places.

12. If $\log_a m = \log_a n$, then $m = n$.

When no base is writ­ten, as in $\log a$, any fixed allowed base may be used (com­monly $10$); laws 9 and 11 hold whichever base it is.

Exam­ple 3#

Eval­u­ate (a) $\log_3 81$ (b) $\log_4 8$ (c) $\log_2 3 \cdot \log_3 8$.

(a) $3^4 = 81$, so $\log_3 81 = 4$.

(b) Write both num­bers as pow­ers of $2$ and use law 5:

$$\displaystyle \log_4 8 = \log_{2^2} 2^3 = \frac{3}{2} \log_2 2 = \frac{3}{2}$$

(c) By the chain law, $\log_2 3 \cdot \log_3 8 = \log_2 8 = 3$.

Exam­ple 4#

Sim­plify $\log_{10} 50 + \log_{10} 2$ and $\log_2 48 - \log_2 3$.

By the prod­uct law, $\log_{10} 50 + \log_{10} 2 = \log_{10} 100 = 2$.

By the quo­tient law, $\log_2 48 - \log_2 3 = \log_2 16 = 4$.

Exam­ple 5#

Show that $\displaystyle \frac{1}{\log_2 6} + \frac{1}{\log_3 6} = 1$.

By law 6, $\displaystyle \frac{1}{\log_2 6} = \log_6 2$ and $\displaystyle \frac{1}{\log_3 6} = \log_6 3$. Adding with the prod­uct law,

$$\log_6 2 + \log_6 3 = \log_6 6 = 1$$

Prac­tice#

1. Write $3^5 = 243$ in log­a­rith­mic form.
2. Write $\log_7 49 = 2$ in expo­nen­tial form.
3. Eval­u­ate $\log_2 64$.
4. Eval­u­ate $\log_{1/2} 8$.
5. Eval­u­ate $\log_9 27$.
6. If $\log_3 (x + 2) = 2$, find $x$.
7. Sim­plify $\log_6 4 + \log_6 9$.
8. Sim­plify $\log_5 75 - \log_5 3$.
9. Eval­u­ate $\log_3 5 \cdot \log_5 9$.
10. Eval­u­ate $5^{\log_5 12}$.
11. Eval­u­ate $\displaystyle \frac{1}{\log_4 12} + \frac{1}{\log_3 12}$.
12. Explain why $\log_{-2} 4$ is not defined.

Answers#