The logarithm

Assume we have an number, say $7$ , and we want to express this number with the base $10$ , that is, we are looking for a number $a$ such that

10^a=7

What is the value of $a$ ? Well, $a$ must be smaller than $1$ , because for $a=1$ we get $10^1=10$ , which is already bigger than $7$ . But $a=0$ is too small, because $10^0=1$ . So $a$ must be somewhere between $0$ and $1$ , but what is the exact value?

Exercise 1

Using the calculator, find a value for $a$ such that

10^a \approx 7

Solution

Trying out some values, we get, for example, that for $a=0.85$ it is

10^{0.85} = 7.07945...

which is close to $7$ .

To find the exact value we need a new tool, called the logarithm. On your calculator, this is the key

\boxed{\texttt{ln log}}

You have to choose the option $\texttt{log}$ (so press the key twice). We will discuss the other option later. So how can we find the value $a$ ? Enter in your calculator

\texttt{log(7)}

The result is

0.8450...

and this is the solution! Indeed, using the calculator we see that

10^{0.8450...}=7

In words, if we want to know

10^\text{what}=7

the answer is the number

\texttt{log(7)}

Of course we do not have to focus on the number $10$ . For example, we could also ask

e^\text{what}=7

and the answer can be found with

\texttt{ln(7)}

which, consulting the calculator, is the number

1.945...

Exercise 2

Determine $a$

$10^a=15$
$e^a=15$

Solution

$a=\texttt{log(15)}=1.176...$
$a=\texttt{ln(15)}=2.708...$

Note that to be more explicit about the base which we want to use to represent the $7$ , we can write $\log_{10}(7)$ rather than $\log(7)$ and $\log_e(7)$ rather than $\ln(7)$ . More generally we have the following definition of the logarithm:

Definition 1

Consider two numbers $b>0$ and $c>0$ . The logarithm base b of c is written as

\log_b(c)

It is

\log_b(c)=a\, \text{ if }\, b^a=c

If this looks complicated, here is a helpful "snails"-diagram: $\log_b(c)=a$ if " $b$ raised to the power of $a$ is $c$ "

For the two bases $e$ (Eulers constant) and $10$ , we do not write the base explicitely:

$\ln(c)=\log_e(c)$ ("natural logarithm")
$\log(c)=\log_{10}(c)$

Example 1

$\log_{10}(1000)=3$ , because $10^3=1000$ (so the $\log$ computers the number of zeros)
$\log_3(9)=2$ because $3^2=9$
$\log_2(0.5)=-1$ because $2^{-1}=0.5$

Before we start with the exercises, a quick note about the names of certain logarithms:

The logarithm base $10$ , that is $\log_{10}$ or $\log$ , is called the common logarithm.
The logarithm base $e$ , that is $\log_e$ or $\ln$ , is called the natural logarithm (presumably because the base $e$ , that is, Euler's constant, occurs in nature a lot).
The logarithm base $2$ , $\log_2$ , is also written as $lb$ , and is called the binary logarithm.

And now some exercises.

Exercise 3

Determine without calculator.
1. $\log(100)$
2. $\log(10)$
3. $\log(1)$
4. $\log(0)$
5. $\log(-1)$
6. $\log(-23)$
7. $\log(0.001)$
8. $\log(10^{-2})$
9. $\log(10^{-1})$
10. $\log(\frac{1}{10})$
11. $\log(\frac{1}{10\,000})$
12. $\log(10^{2.37})$
13. $\log(10^{3x})$
14. $\log(10^{3x^4+3})$
15. $\log(4+6)$
16. $10^{\log(4)}$
Determine without calculator.
1. $\log_{3}(9)$
2. $\log_{5}(\frac{1}{25})$
3. $\log_{2}(1024)$
4. $\log_{2}(0.125)$
5. $\log_{100}(10)$
6. $\log_{3}(\frac{1}{81})$
7. $\log_{3}(\sqrt[10]{9})$
8. $\log_{3}(\frac{1}{\sqrt[5]{27}})$
9. $\log_{\pi}(\frac{1}{\pi})$
10. $\ln(e^{1/7})$
11. $e^{\ln(6)}$
Can you simplify the expression?
1. $\log_{b}(0)$
2. $\log_{b}(1)$
3. $\log_{b}(b)$
4. $\log_{b}(\sqrt{b})$
5. $\log_{b}(\frac{1}{b})$
6. $\log_{b}(\frac{1}{\sqrt{b}})$
7. $\log_{b}(b^{101})$
8. $\log_{b}(-1)$
9. $\log_{b}(b^{2x-1})$
10. $b^{\log_b(4)}$

Solution

Find the logarithm
1. $\log(100)=2$ because $10^2=100$
2. $\log(10)=1$ because $10^1=10$
3. $\log(1)=0$ because $10^0=1$
4. $\log(0)$ does not exist, because there is no number $a$ with $10^a=0$
5. $\log(-1)$ does not exist, because there is no number $a$ with $10^a=-1$
6. $\log(-23)$ does not exist, because there is no number $a$ with $10^a=-2$
7. $\log(0.001)=\log(10^{-3})=-3$
8. $\log(10^{-2})=-2$
9. $\log(10^{-1})=-1$
10. $\log(\frac{1}{10})=\log(10^{-1})=-1$
11. $\log(\frac{1}{10\,000})=\log(10^{-4})=-4$
12. $\log(10^{2.37})=2.37$
13. $\log(10^{3x})=3x$
14. $\log(10^{3x^4+3})=3x^4+3$
15. $\log(4+6)=\log(10)=1$
16. $10^{\log(4)}=4$ because by definition $\log(4)$ is the number $a$ with $10^a=4$ .
Determine without calculator.
1. $\log_{3}(9)=2$ because $3^2=9$
2. $\log_{5}(\frac{1}{25})=\log_5(5^{-2})=-2$
3. $\log_{2}(1024)=\log_{2}(2^{10})=10$
4. $\log_{2}(0.125)=\log_2(\frac{1}{8})=\log_2(2^{-3})=-3$
5. $\log_{100}(10)=\log(100^{0.5})=0.5$
6. $\log_{3}(\frac{1}{81})=\log_3(3^{-4})=-4$
7. $\log_{3}(\sqrt[10]{9})=\log_3(9^{1/10})=\log_3(3^{2/10})=0.2$
8. $\log_{3}(\frac{1}{\sqrt[5]{27}})=\log_3(3^{-3/5})=-0.6$
9. $\log_{\pi}(\frac{1}{\pi})=\log_{\pi}(\pi^{-1})=-1$
10. $\ln(e^{1/7})=\frac{1}{7}$
11. $e^{\ln(6)}=6$ because by definition $\ln(6)$ is the number $a$ with $e^a=6$ .
We always assume $b>0$ :
1. $\log_{b}(0)$ no, for any $b>0$ this number does not exist
2. $\log_{b}(1)=0$ because $b^0=1$ for any $b>0$
3. $\log_{b}(b)=1$ because $b^1=b$ for any $b>0$
4. $\log_{b}(\sqrt{b})=0.5$ because $b^{0.5}=\sqrt{b}$ for any $b>0$
5. $\log_{b}(\frac{1}{b})=-1$ because $b^{-1}=\frac{1}{b}$ for any $b>0$
6. $\log_{b}(\frac{1}{\sqrt{b}})=-\frac{1}{2}$ because $b^{-1/2}=\frac{1}{\sqrt{b}}$ for any $b>0$
7. $\log_{b}(b^{101})=101$ because $b^{101}=b^{101}$ for any $b>0$
8. $\log_{b}(-1)$ does not exist, because there is no $a$ with $b^a=-1$
9. $\log_{b}(b^{2x-1})=2x-1$ because $b^{2x-1}=b^{2x-1}$
10. $b^{\log_b(4)}=4$ because by definition for $a=\log_b(4)$ it is $b^a=4$ .

Let us summarise some useful properties which follow from the exercise above:

Theorem 1

For every base $b>0$ the following is true: 0. $\log_b(c)$ does not exist for $c\leq 0$

$\log_b(1)=0$
$\log_b(b)=1$
$\log_b(b^u)=u$ for every number $u$
$b^{\log_b(c)}=c$ for every $c>0$

Proof

Here are the proofs:

There is no number $a$ with $b^a=0$ or $b^a$ is negative (for $b>0$ ).
$\log_b(1)=0$ because $b^0=1$
$\log_b(b)=1$ because $b^1=b$
$\log_b(b^n)=n$ because $b^n=b^n$
By definition, for the number $a=\log_b(c)$ it is $b^a=c$ , so indeed $b^{\log_b(c)}=c$

Exercise 4

Argue, why is $\log_b(c\cdot d)=\log_b(c)+\log_b(d)$ ?
Is the following true? $\log_5(3x)=\log_5(3)+\log_5(x)$ for all $x$ ?
Is the following true? $\log_5(3+x)=\log_5(3)\cdot \log_5(x)$ for all $x$ ?

Solution

To show that $\log_b(c\cdot d)=\log_b(c)+\log_b(d)$ is correct, the equation must also be correct if we raise $b$ to the power of the left side and the right side of the equation: $\underbrace{b^{\log_b(c\cdot d)}}_{=c\cdot d}=b^{\log_b(c)+\log_b(d)}$ Using the power laws, we get for the right side $\begin{array}{lll} b^{\log_b(c)+\log_b(d)}&=&b^{\log_b(c)}\cdot b^{\log_b(d)}\\ &=& c\cdot d \end{array}$ and we see that the left side is the same as the right side. This concludes the proof.
It is correct (see question 1).
No, it is wrong. Take for example $x=1$ . For the left side we get $\log_5(3+x)=\log_5(3+1)=\log_5(4)\neq 0$ and for the right side we get $\log_5(3)\cdot \log_5(x) = \log_5(3)\cdot \underbrace{\log_5(1)}_{=0}=0$