Logarithms Lecture 6: Other bases

So far, we have been saying that logarithms “count zeroes”. For example,

log (10,000) = 4

because 10,000 has four zeroes. We have been extrapolating from there.

A “zero” really means a factor of 10; the number 10,000 has four zeroes because

10,000 = 10 x 10 x 10 x 10,

which we can also write as $10^4$ . But what is special about 10?

The reason to choose 10 is because it works well with the way we write numbers. (See the next optional lecture for more on that.) But mathematically, there is nothing distinguished about 10. In principle, we could use any positive number in place of 10 in all of our reasoning so far.

In practice, the numbers other than 10 in common usage are the number 2, and the number e, which is approximately 2.71828. The number we use in place of 10 is called the “base” of the logarithm. The log base 2 is written $\log_2$ ; the log base b is written $\log_b$ . One exception: usually for $\log_e$ , people instead write $\ln$ , which stands for the “natural” logarithm. The default assumption is 10, so if we write just $\log$ it means $\log_{10}$ ; sometimes people write $\log_{10}$ to avoid ambiguity. I’ll explain some reasons one might use 2 at the end of this lecture; I’ll explain the use of the strange number e later on.

Logarithms with a base of 2

So, rather than counting the tens in a number, we could have started this whole story by counting the twos instead. We would start off with numbers that are a whole number of twos:

$\log_2 2 = 1$
$\log_2 4 = 2$
$\log_2 8 = 3$
$\log_2 16 = 4$
$\log_2 32 = 5$
$\log_2 64 = 6$
$\log_2 128 = 7$
$\log_2 256 = 8$
and so forth.

As before, multiplying two numbers has the effect of adding the amount of twos in the two numbers. For example, 8 is 3 twos, and 64 is 6 twos, so 8 x 32 = 256, which is 9 twos total (that is, (2 x 2 x 2) x (2 x 2 x 2 x 2 x 2 x 2) = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2).

We could then figure out what a “half a two” ought to mean. Following the same reasoning as we did before, since we are working multiplicatively, a “half a two” would be a number that multiplies by itself to give 2. That number would be $\sqrt{2}$ , which is approximately 1.414, or more roughly, about 1.4. So a half of a two is about 1.4. As we did before, we could use that to estimate some other logs. For example, since 90 is approximately 64 x 1.4,

$\log_2 90 \simeq \log_2 (64 \times 1.4) = \log_2 64 + \log_2 1.4\simeq 6+0.5 = 6.5$ .

We could then go on to figure out how to estimate logs base 2, and figure out logs base 2 for numbers smaller than 1, all as before.

Problem 1: In a similar way to how we did it before, try to find the logarithms base 2 of the following numbers. (Exactly if you can do it, otherwise approximately.)
a. 512
b. 0.5
c. 0.25
d. 1
e. 3
f. 5

Approximate relation to base 10 logarithms

We have done a lot of work on logarithms base 10 already, so it would be nice not to have to start over again from the beginning. As it turns out, the log base 2 has a fairly simple relationship to the log base 10, which we can use to convert logs base 2 into logs base 10. In this way, we can avoid doing the work again: when we want a log base 2, we can find the log base 10, and then convert.

So, how does the relationship work?

First I’ll show you how it works approximately, for mental calculations. Then I’ll go back and do it exactly.

Remember that we have established that $\log_{10}2\simeq 0.3$ . (Right? Review it now if you need to. I’ll wait. … OK, you’re back? Great.) That is to say, a two is worth about 0.3 of a zero. More mathematically stated,

$2\simeq 10^{0.3}$ .

Let’s turn that around: how many twos do I need to make up one zero? (That is, to make up one factor of ten.) It’s a bit more than 3, because

$2 \times 2 \times 2 = 8 < 10$ ,

or, stated differently,

$2 \times 2 \times 2 \simeq 10^{0.3} \times 10^{0.3} \times 10^{0.3}= 10^{0.9}$ ,

which comes up shy of $10^1=10$ . We need an extra 0.1 of a zero, that is, an extra $10^{0.1}$ . But that is one third of a two, because

$10^{0.1} \times 10^{0.1} \times 10^{0.1}=10^{0.3}\simeq 2$ .

So, in summary, one zero is worth about three and one third twos:

$2^{3.33}\simeq 10$ .

This is all we need! Because if we know how to find the logarithm base 10 of a number, that logarithm represents how many zeroes, i.e. tens, are in the number. Then, the number of twos in our number is just the number of tens in it, times 3 and a third (since every ten is 3 and a third twos).

For example, 10,000 has four zeroes: 10,000 = 10 x 10 x 10 x 10. But each of those tens is 3.33 twos:

$10,000 = 10\times 10\times 10\times 10 = 2^{3.33}\times 2^{3.33}\times 2^{3.33}\times 2^{3.33}$
$= 3.33 + 3.33 + 3.33 + 3.33 = 3.33\times 4=13.33$ ,

so $\log_2(10,000)\simeq 13.33$ .

All I’m doing is taking the number of tens in a number, and multiplying it by 3.33 to get the number of twos in that number. The rule is simply

$\log_2(x) \simeq (3.33)\log_{10}(x)$

for any number x.

For mental calculations, it is usually best to remember the factor of 3.33 as being $3\frac{1}{3}$ : you take the log base 10 of the number, multiply it by 3, and then add one third of that number. For example, to find $\log_2(350)$ , I would first find $\log_{10}(350)$ , which is about 2.55 (remember why?? Do it now!). Then I would calculate:

$\log_2(350)\simeq 2.55 + 2.55 + 2.55 + \text{one third of}\ 2.55$
$=7.65 + \text{about}\ 0.85$
$=8.5$ ,

so $\log_2(350)$ is about 8.5. (I found one third of 2.55 by dividing 3 into 2.55.)

Problem 2: Find the logarithm of each number base 2 approximately. Do it mentally, by finding the logarithm base 10 (as you’ve learned to do before), and multiplying the answer by 3 and a third.
a. 100
b. 1024 (You can find this exactly; do the approximate method as well as a check)
c. 670,000
d. 0.00045

Problem 3: Recall the graph you drew for y = log x (for base 10 logs). Suppose you were to now draw the graph for $y = \log_2 x$ . Given the relationship between the two types of logs, how would the graphs be related?

Exact relation to base 10 logarithms

To find the exact relationship between base 2 and base 10 logarithms, I just need to go back over the argument I just made, and replace all my approximate statements with exact ones.

I started out by saying that, since $\log_{10} 2$ is about 0.3, this means that a two is worth about 0.3 of a ten:

$10^{0.3} = 2$ .

I could have said that without the approximation:

$10^{\log_{10} 2} = 2$ .

(Say what this means in words to yourself a few times until it starts to make sense!!)

Next, I wanted to turn it around and figure out how many twos I needed to make up one ten. I said

$10^{0.3}\times 10^{0.3}\times 10^{0.3}=10^{0.9}$

comes up short. Well, since multiplying the numbers adds the number of tens (exponents), I’m trying to figure out

$0.3 \times \mathrm{what??} = 1$

so the answer is

$\mathrm{what??} = \dfrac{1}{0.3}$ ,

which is how I got approximately 3 and a third. But that 0.3 is just an approximation for $\log_{10} 2$ , so the exact value is

$\mathrm{what??} = \dfrac{1}{\log_{10}2}$

What, exact value of what? Remember we were trying to find how many twos I need to make up one ten. The approximate answer is 3.33; the exact answer is

$\dfrac{1}{\log_{10}2}$

twos count as one ten. That is,

$2^{\left(\dfrac{1}{\log_{10}2}\right)} = 10$ .

This in turn means that, if I count the tens in a number, then I have to multiply this by $\dfrac{1}{\log_{10}2}$ in order to find out how many twos there are:

$\log_2(x) = \dfrac{1}{\log_{10}2}\times\log_{10}(x)$ .

The log base 2 of a number is just a constant value times the log base 10 of that number. The constant is approximately 3.33, and it is exactly $\dfrac{1}{\log_{10}2}$ .

Bases other than 2

If we were to start counting the number of threes in a number, the whole discussion above would go pretty much the same. We would be computing the “logarithm base 3”. We would end up with the relation

$\log_3(x) = \dfrac{1}{\log_{10}3}\times\log_{10}(x)$ ;

if we work out the numerical factor in front on the right side, we would get

$\log_3(x) \simeq (2.10) \times\log_{10}(x)$ .

Similarly, if b is any positive number, the log base b is related to log base 10 by

$\log_b(x) = \dfrac{1}{\log_{10}b}\times\log_{10}(x)$ .

A handy case to remember is the conversion for natural logarithms, base e, which we’ll talk more about a bit later. The rule is

$\ln (x) = \log_e(x) = \dfrac{1}{\log_{10}e}\times\log_{10}(x)$ ;

numerically, this works out as

$\ln (x) \simeq (2.30)\log_{10}(x)$ ,

which is quite useful for computing natural logs mentally. For example, if I want to find ln(3700), I first mentally estimate log(3700)=3.57 (right?? check my math!), and then multiply by 2.30. Here’s my method for multiplying mentally by 2.3:

First multiply by 2. I know 3.50 x 2 = 7; then, 0.07 x 2 = 0.14.
So 3.57 x 2 = (3.50 x 2) + (0.07 x 2) = 7 + 0.14 = 7.14.
Then, multiply by 3: since I’ve already multiplied 3.55 by 2, I just add one more:
3.55 x 3 = (3.55 x 2) + (3.55 x 1) = 7.14 + 3.55 = 10.69.
Now move the decimal point:
3.55 x 0.30 = 1.069, or roughly 1.07.
Finally, I add 3.55 x 2 plus 3.55 x 0.30:
3.55 x 2.30 = (3.55 x 2) + (3.55 x 0.30) = 7.14 + 1.07 = 8.21.
So, my estimate is ln(3700) is about 8.21.

I’m writing out many steps, but with a little practice this isn’t hard to do mentally (or at least with pencil and paper). Now, there is an ln key on the calculator, so I can check this! I enter ln(3700), and get:

Ha ha, that’s just my cat. Let me try again:

That’s better! My mental estimate was 8.21, so things are going pretty well.

Problem 4: Compute the natural log of each of these numbers approximately. Do it mentally by finding the log of the number base 10, and then multiplying by 2.30. Check each answer on a calculator.
a. 10
b. 1,000,000
c. 73,368
d. 0.002

Applications of logs with different bases

Why would we use bases other than 10?

Base 2 is useful in applications where it is natural to talk about doubling or halving things. In population growth, cell growth, epidemiology, and other similar growth problems, exponential growth is typical in early stages, and people like to express things in terms of “doubling time”. Therefore log base 2 is helpful. In problems of radioactive decay (like carbon dating), the decay is exponential, and people like to talk about the rate in terms of “half-life”, so, again, log base 2 is helpful. In music, doubling the frequency (multiplicatively) of a tone increases the note (additively) by an octave. So logs are involved, and in particular log base 2 is the sensible one, since doubling a frequency is the most important relationship between notes harmonically.

It will be easier to give more detail about application of $\log_{10}$ and $\log_2$ once I tell you a bit more about logs and equations, in “Logarithms Version 3” that is coming up soon. Once we have that, I’ll give you some more detail for some of them.

Logarithms base e, “natural” logs, appear in all sorts of places, particularly in physics and in probability, but many other surprising places too. The reason for using such a strange base, e=2.71828182845904523536028747135…, depends on calculus. The calculus story also gives a whole different way of motivating logarithms. To get the full story, you’ll need to take calculus here! But I will try to at least give you the beginning of the story, towards the end of these lectures.

Optional Appendix: What is special about 10?

You can skip this section if you like. It isn’t needed in anything after this.

Why DO we start with logarithms base 10? It’s particularly easy to compute logarithms in base 10: if you see the number 12,760,368, you know right away that its logarithm base 10 is between 7 and 8, and a little thought can give you a more accurate estimate. What is the logarithm of 12,760,368 base 2? That’s much harder to see immediately; you would have to know your powers of 2 up that high to see it directly. The easiest way to find the base 2 logarithm of a number like that is to find the base 10 logarithm and convert.

But why should 10 be easier? It’s because our numbering system is based on tens.

There is nothing mathematically special about 10. Ten is a historical accident: some 500 million years ago (biologists can correct my date!), there were a bunch of different body plans in competition, and the one which one and led to vertebrate life happened to have five things on the end of each appendage, rather than some other number.

Consequently, we group things every ten. And then once we have ten groups of ten, we group that too, so we count in groups of 10 x 10 = 100, and so on. The number 3,148 means 8 single things, 4 groups of 10, 1 group of 10 groups of 10, and 3 groups of 10 groups of 10 groups 0f 10:

3,148 = 3 x (10 x 10 x 10) + 1 x (10 x 10) + 4 x (10) + 8

Actually, even with our fingers, using 10 is a historical accident: at various times, people have also grouped in groups of 20, 60, and 12. In modern times, people have used groups of 2, 8, and 16, for computer reasons. If the Mayan society had been more long-lived, we might be counting in groups of 20 today.

For example, if we used 8 as the basis of our number system, we would count

1, 2, 3, 4, 5, 6, 7, 10, 11, 12, 13, 14, 15, 16, 17, 20, 21, …

In this system, “10” doesn’t mean ten; it means eight: 1 group of eight, 0 singles. The number “37” in base 8 means 31 in our system: “37” means 3 x 8 + 7 x 1. The number “2457” means

“2457” = 2 x (8 x 8 x 8 ) + 4 x (8 x 8 ) + 5 x (8) + 7

Base 8 only has 8 symbols: 0, 1, 2, 3, 4, 5, 6, 7. There is no need for a symbol for 8 or 9.

If we used a base bigger than 10, we would need more symbols. In base 16, the standard way people count is

1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, 10, 11, 12, … , 1E, 1F, 20, 21, ….

The letter “A” stands for 10, “F” stands for 15. For example, the number “D4” stands for 13 groups of 16 plus 4 singles (so equals 212 in decimal notation).

Let’s try base 2 (also called “binary”). We would count (decimal on left, binary on right):

1 = “1”
2 = “10”
3 = “11”
4 = “100”
5 = “101”
6 = “110”
7 = “111”
8 = “1000”
9 = “1001”
etc.

So, for example,

$\text{``}1100110\text{''}= 1 \times 2^6 + 1 \times 2^5 + 0\times 2^4 + 0\times 2^3 + 1\times 2^2 + 1\times 2^1 + 0\times 2^0$
$=1 \times 64 + 1 \times 32 + 0\times 16 + 0\times 8 + 1\times 4+ 1\times 2 + 0\times 1$
$=102$ .

If this seems strange, it’s just because we aren’t used to it; if we were used to counting in base 12, say, then base 10 would seem just as strange and complicated. There is nothing mathematically special about 10.

How is this related to logarithms? Well, check this out. The numbers I am writing in quotes are in binary; numbers without quotes are ordinary decimal notation:

$\log_2(\text{``}10\text{''}) = \log_2(2) = 1$
$\log_2(\text{``}100\text{''}) = \log_2(4) = 2$
$\log_2(\text{``}1000\text{''}) = \log_2(8) = 3$
$\log_2(\text{``}10000\text{''}) = \log_2(16) = 4$
$\log_2(\text{``}100000\text{''}) = \log_2(32) = 5$
$\log_2(\text{``}1000000\text{''}) = \log_2(64) = 6$
$\log_2(\text{``}10000000\text{''}) = \log_2(128) = 7$
…

The logarithm base 2 IS counting the zeroes—but it’s counting the zeroes when the number is written in base 2!

So if I write my numbers in base 2 notation, then I could do the whole development that I did from the beginning: $\log_2$ counts the zeroes, then figure out how to interpolate, etc etc. Logs with any whole number base b can be thought of as counting the zeroes, when you are using the number b as the basis for your counting system. (But this isn’t going to work quite as nicely for logs base e!!)