The word “logarithm” makes most people think about some mythical mathematical beast. Surely logarithms are many headed, breathe fire and are extremely difficult to understand. Nonsense! Logarithms are simple. It will take you at most a couple of pages to get used to manipulating them, and that is a good thing because logarithms are used all over the place.

For example, the strength of your sound system is measured in logarithmic units called decibels $[\textrm{dB}]$. This is because your ear is sensitive only to exponential differences in sound intensity. Logarithms allow us to compare very large numbers and very small numbers on the same scale. If we were measuring sound in linear units instead of logarithmic units then your sound system volume control would have to go from $1$ to $1048576$. That would be weird no? This is why we use the logarithmic scale for the volume notches. Using a logarithmic scale, we can go from sound intensity level $1$ to sound intensity level $1048576$ in 20 “progressive” steps. Assume each notch doubles the sound intensity instead of increasing it by a fixed amount, the first notch corresponds to $2$, the second notch is $4$ (still probably inaudible) but by the time you get to sixth notch you are at $2^6=64$ sound intensity (audible music). The tenth notch corresponds to sound intensity $2^{10}=1024$ (medium strength sound) and the finally the twentieth notch will be max power $2^{20}=1048576$ (at this point the neighbours will come knocking to complain).

You are probably familiar with these concepts already:

• $b^x$: the exponential function base $b$
• $\exp(x)=e^x$: the exponential function base $e$, Euler's number
• $2^x$: exponential function base $2$
• $f(x)$: the notion of a function $f:\mathbb{R}\to\mathbb{R}$
• $f^{-1}(x)$: the inverse function of $f(x)$. It is defined in terms of

$f(x)$ such that the following holds $f^{-1}(f(x))=x$, i.e.,

  if you apply $f$ to some number and get the output $y$,
  and then you pass $y$ through $f^{-1}$ the output will be $x$ again.
  The inverse function $f^{-1}$ undoes the effects of the function $f$.

NOINDENT In this section we will play with the following new concepts:

• $\log_b(x)$: logarithm of $x$ base $b$. This is the inverse function of $b^x$
• $\ln(x)$; the “natural” logarithm base $e$. This is the inverse of $e^x$
• $\log_2(x)$: the logarithm base $2$ is is the inverse of $2^x$

I say play, because there is nothing much new to learn here: logarithms are just a clever way to talk about the size of number – i.e., how many digits the number has.

The main thing to realize is that $\log$s don't really exist on their own. They are defined as the inverses of the corresponding exponential function. The following statements are equivalent: \[ \log_b(x)=m \ \ \ \ \ \Leftrightarrow \ \ \ \ \ b^m=x. \]

For logarithms with base $e$ one writes $\ln(x)$ for “logarithme naturel” because $e$ is the “natural” base. Another special base is $10$ because we use the decimal system for our numbers. $\log_{10}(x)$ tells you roughly the size of the number $x$—how many digits the number has.

When someone working for the system (say someone with a high paying job in the financial sector) boasts about his or her “six-figure” salary, they are really talking about the $\log$ of how much money they make. The “number of figures” $N_S$ in you salary is calculated as one plus the logarithm base ten of your salary $S$. The formula is \[ N_S = 1 + \log_{10}(S). \] So a salary of $S=100\:000$ corresponds to $N_S=1+\log_{10}(100\:000)=1+5=6$ figures. What will be the smallest “seven figure” salary? We have to solve for $S$ given $N_S=7$ in the formula. We get $7 = 1+\log_{10}(S)$ which means that $6=\log_{10}(S)$ and using the inverse relationship between logarithm base ten and exponentiation base ten we find that $S=10^6 = 1\:000\:000$. One million per year. Yes, for this kind of money I see how someone might want to work for the system. But I don't think most system pawns ever make it to the seven figure level. Even at the higher ranks, the salaries are more in the $1+\log_{10}(250\:000) = 1+5.397=6.397$ digits range. There you have it. Some of the smartest people out there selling their brains out to the finance sector for some lousy $0.397$ extra digits. What wankers! And who said you need to have a six digit salary in the first place? Why not make $1+\log_{10}(44\:000)=5.64$ digits as a teacher and do something with your life that actually matters?

Let us now discuss two important properties that you will need to use when dealing with logarithms. Pay attention because the arithmetic rules for logarithms are very different from the usual rules for numbers. Intuitively, you can think of logarithms as a convenient of referring to the exponents of numbers. The following properties are the logarithmic analogues of the properties of exponents

The first property states that the sum of two logarithms is equal to the logarithm of the product of the arguments: \[ \log(x)+\log(y)=\log(xy). \] From this property, we can derive two other useful ones: \[ \log(x^k)=k\log(x), \] and \[ \log(x)-\log(y)=\log\left(\frac{x}{y}\right). \]

Proof: For all three equations above we have to show that the expression on the left is equal to the expression on the right. We have only been acquainted with logarithms for a very short time, so we don't know each other that well. In fact, the only thing we know about $\log$s is the inverse relationship with the exponential function. So the only way to prove this property is to use this relationship.

The following statement is true for any base $b$: \[ b^m b^n = b^{m+n}, \] which follows from first principles. Exponentiation means multiplying together the base many times. If you count the total number of $b$s on the left side you will see that there is a total of $m+n$ of them, which is what we have on the right.

If you define some new variables $x$ and $y$ such that $b^m=x$ and $b^n=y$ then the above equation will read \[ xy = b^{m+n}, \] if you take the logarithm of both sides you get \[ \log_b(xy) = \log_b\left( b^{m+n} \right) = m + n = \log_b(x) + \log_b(y). \] In the last step we used the definition of the $\log$ function again which states that $b^m=x \ \ \Leftrightarrow \ \ m=\log_b(x)$ and $b^n=y \ \ \Leftrightarrow \ \ n=\log_b(y)$.

We will now discuss the rule for changing from one base to another. Is a relation between $\log_{10}(S)$ and $\log_2(S)$?

There is. We can express the logarithm in any base $B$ in terms of a ratio of logarithms in another base $b$. The general formula is: \[ \log_{B}(x) = \frac{\log_b(x)}{\log_b(B)}. \]

This means that: \[ \log_{10}(S) =\frac{\log_{10}(S)}{1} =\frac{\log_{10}(S)}{\log_{10}(10)} = \frac{\log_{2}(S)}{\log_{2}(10)}=\frac{\ln(S)}{\ln(10)}. \]

This property is very useful in case when you want to compute $\log_{7}$, but your calculator only gives you $\log_{10}$. You can simulate $\log_7(x)$ by computing $\log_{10}(x)$ and dividing by $\log_{10}(7)$.