Who understands X?

[Eclectic Mystic]

I'll start:

Who understands what logarithms are?

OK, go!

[Bishadi]

Originally Posted by Eclectic Mystic

I'll start:

Who understands what logarithms are?

OK, go!

The logarithm is perhaps the single, most useful arithmetic concept in all the sciences; and an understanding of them is essential to an understanding of many scientific ideas. Logarithms may be defined and introduced in several different ways. But for our purposes, let's adopt a simple approach. This approach originally arose out of a desire to simplify multiplication and division to the level of addition and subtraction. Of course, in this era of the cheap hand calculator, this is not necessary anymore but it still serves as a useful way to introduce logarithms. The question is, therefore:

Is there any operation in mathematics which produces a multiplication by the performance of an addition?
With not too much thought, the answer should come to you.
What is 23 x 24.
The answer is 2 7 which is obtained by adding the powers 3 and 4. This is correct, of course, since 23 x 24 is just seven 2s multiplied together. Note that this addition trick does not work for the case of 33 x 24. The base numbers must be the same, as in the first case, where we used 2.
In general, this addition trick can be written as pa x pb = pa+b. This expression will do our job of multiplying any two numbers, say 1.3 and 6.9, if we can only express 1.3 as pa and 6.9 as pb.
What number will we use for the base p? Any number will do, but traditionally, only two are in common use:
Ten (10) and the transcendental number e (= 2.71828...), giving logarithms to the base 10 or common logarithms (log), and logarithms to the base e or natural logarithms (ln).
If you would like to know why this strange number e is used click here.
Let's first talk about logarithms to the base 10 or common logs. We thus choose to let our number 1.3 be equal to 10a.
1.3 = 10a
`a' is called "the logarithm of 1.3". How large is `a'? Well, it's not 0 since 100 = 1 and it's less than 1 since 101 = 10. Therefore, we see that all numbers between 1 and 10 have logarithms between 0 and 1. If you look at the table below you'll see a summary of this. Number range
1 - 10 or 100 - 101
10 - 100 or 101 - 102
100 -1000 or 102 - 103
etc.Logarithm Range
0 -1
1 - 2
2 - 3
etc.
You see, we have the number range listed on the left and the logarithm range listed on the right. For numbers between 1 and 10, that is between 100 and 101, the logarithm lies in the range 0 to 1. For numbers between 10 and 100, that is between 101 and 102, the logarithm lies in the range 1 to 2, and so on. Now in the bad old days before calculators, you would have to learn to use a set of logarithm tables to find the logarithm of our number, 1.3, that we asked for earlier. But nowadays, you can get it at the press of a button on your calculator.

I'm going to take a moment to discuss your calculator. If you don't have a calculator with scientific functions on it, you should get one before proceeding in this tutorial since the rest depends on it.
Most calculators are very straightforward in obtaining the logarithm. They either have two logarithm keys or a dual function key. In any case, the labels will be `log' and `ln' which is often pronounced `lon'.
Log is the key for logs to the base 10 and ln is for natural logs. We want logs to the base 10 in our example so we use `log'. Enter 1.3 on your calculator, and then press the log key.
Do you have 0.113943? You should have. This number then is `a' back in our previous expression and therefore the logarithm of 1.3. Pause now and determine `b' in that expression, the logarithm of 6.9.
You should have 0.838849 for the log of 6.9. If not, review what we have done and try again. Now we are going to do something silly in view of the fact that you have a calculator. We're going to use the two logarithms you have evaluated to find the product of 1.3 and 6.9. Of course, you can do it quickly with your calculator, but this will show that logarithms do what they are supposed to do. According to our original idea, the sum of the two logarithms was supposed to be the logarithm of the answer.
Now add the two logarithms. The sum is 0.952792. This is the logarithm of the answer. If we only knew what number had 0.952792 as its logarithm, we would know the value of 1.3 x 6.9.
The problem of finding a number when you know its logarithm is called finding the "antilogarithm" or sometimes "exponentiation". Again, lets look at your calculator. Here is where calculators differ a lot and I hope I mention one that is something like yours.
You should look for a key on your calculator that says something like 10x or 10y. If so, then pressing that key will take the antilog of the number in the display. Alternatively, your calculator may have an "inverse" key. If so, then pressing inverse and then log will take the antilog of the number in the display. Enter 0.952792 into your calculator and find the antilog.
Did you get 8.97? If not, try again. Of course, you might have got something like 8.96999 but of course that really is 8.97. Now, multiply 1.3 x 6.9 on your calculator and you'll see that 8.97 is indeed the correct answer.
The whole operation could be done with natural logarithms as well as shown below.
1.3 x 6.9 = ?
ln 1.3 = 0.262364
ln 6.9 = 1.931521
total = 2.193885 i.e. 1.3 = e0.262364
i.e. 6.9 = e1.931521
i.e. 1.3 x 6.9 = e2.193885


antiln 2.193885 = 8.97
If the sum of logarithms gives the product of two numbers, then the difference gives the quotient.
In the table below, I've taken the difference between the ln of 1.3 and the ln of 6.9. Check it on your calculator.
1.3/6.9 = ?
ln 1.3 = 0.262364
ln 6.9 = 1.931521
ln 1.3/6.9 = -1.669157
antiln (-1.669157) = 0.1884Don't be afraid of the negative sign. It simply means that the answer is less than 1. Enter -1.669157 on your calculator, then find its antiln. Note we are working with natural logs in this example.

If you didn't get 0.1884, try again. Of course, this is just 1.3 divided by 6.9. In the table below, I have done the whole problem over again using common logs. Pause here and check it. log 1.3 = 0.113943
log 6.9 = 0.838849
log(1.3/6.9) = -0.724906
antilog (-0.724906) = 0.1884
The logarithmic and exponential functions are very important since many physical and biological processes can be described by them.
For example, suppose you have a certain number of radioactive atoms at time t = 0. Let's let this number be N0. Radioactivity behaves in such a way that the number N of radioactive atoms remaining at a later time t is given by a linear variation of the logarithm of N with t. That is, a graph of ln N vs t is a straight line. You know that the equation of such a straight line is given by y = mx+b where m is the slope and b is the y intercept. Therefore, the equation of radioactivity is ln N = -kt + ln N0 where ln N0 is the y intercept and the slope of the line is -k.

Let's now examine the equation we derived for radioactivity, ln N = ln N0 -kt. Here is where it is important to be able to do algebra with logarithms. Lets get the logarithms on one side so that we get
ln N - ln N0 = -kt.
But we know that the difference of logarithms is the logarithm of the quotient so the left-hand side becomes ln N/N0. Now let's take antilogarithms and do the right side first. The antiln of any quantity is the number e to the power of that quantity so the right-hand side becomes e-kt. The left-hand side is the antiln of the ln and so it just becomes N divided by N0. Finally, we can rearrange to put the final equation in the form

N = N0 e-kt
which is called the equation of "exponential decay", so you can see why taking an antiln is often called "exponentiation".
Quiz Try the following questions:
1. ln (7.42) = ?
2. log (7.42) = ?
3. ln (ekt) = ?
4. log (ekt) = ?
5. antilog 0.8704 = ?
6. antiln 2.0042 = ?
7. 100.8704 = ?
8. e2.0042 = ?






Now do your quiz........

[StarshipEnterprise]

@ Bishadi:

*cough* plagiarism *cough*

@ Eclectic Mystic:

Care to explain the rules of this game? And why the heck is this in abrahamic religions???

[Dream]

It looks like a sweet explanation of logarithms to me. I don't follow what it is saying about 23 x 24, because that is not 2^7 power. 2^7=128 whereas 23 x 24 = 552. Also, 23 is not divisible by 2 and 23 x 24 = 2^9.1085 power.

[wil]

Originally Posted by StarshipEnterprise

Care to explain the rules of this game? And why the heck is this in abrahamic religions???

Who understands what forum sections are?

(I went ahead and moved to the lounge)

Originally Posted by Eclectic Mystic

I'll start:

Who understands what logarithms are?

OK, go!

Who understands grokking?

next!

[dauer]

Grok is a word that comes from Robert Heinlein's book, Stranger in a Strange Land. He wanted to use a word that was foreign and could simulate the biblical concept of "knowing."

A person can grok a concept. A person can also grok another person.

[kiwimac]

Waiting for fullness!

[StarshipEnterprise]

Originally Posted by Dream

It looks like a sweet explanation of logarithms to me. I don't follow what it is saying about 23 x 24, because that is not 2^7 power. 2^7=128 whereas 23 x 24 = 552. Also, 23 is not divisible by 2 and 23 x 24 = 2^9.1085 power.

by 23 and 24 he meant 2^3 and 2^4. Copy and paste can sometimes mess up your format

[Dream]

I was just testing you. Well done. Well done. Splendid.

[Eclectic Mystic]

Wait a second! Anyone who looks into this will realize this was originally placed in the appropriate forum.

[wil]

Originally Posted by Eclectic Mystic

Wait a second! Anyone who looks into this will realize this was originally placed in the appropriate forum.

Ok I will move it back if warrented, explain yourself.

I think most of us are lost.

We need to know what the title means, vs. the original post, vs. Ok, go.

Title could be Christian, but Abrahamic??

OP looks like either a game or a mathematical question.

[juantoo3]

Originally Posted by dauer

use a word that was foreign and could simulate the biblical concept of "knowing."

As in, "knowing" in the Biblical sense?

Not to seem impudent, but could a person then correctly say "grok you?"

Not that I am saying this to you, but trying to understand. And enjoying a bit of tongue in cheek fun playing with words.

[dauer]

juan,

I don't think so.

-- Dauer

[juantoo3]

Who understands Y?

Y is the sky blue?

Y is the sea green?

Y do we even bother?

Y knot?

[bob x]

I understand X better than anybody, of course.