# Calculating logarithms

#### Calculating logarithms

So I've been struggling with working out how logarithms work and have finally decided that I'm tired of that. As far as I understand the basic gist of it a logarithms is effectively the analog of division for exponentials. The problem comes in with most descriptions I've found which go something along the lines of "take the logarithm of x, add it to y and the do something with the antilogarithm and you have the answer". Yeah, great, and how do I calculate the logarithm and the antilogarithm exactly?

So, sarcasm aside and ignoring the issues of /using/ logarithms for a moment. How do I calculate the log and anti-log for... something. Also writing that sentence reminds me, what are logs and anti-logs calculated "against".

I truly apologise for how vague and terribly worded my question is, but it's a direct reflection of how bad my grasp is on this topic. And this is as far as I'd gotten in all the reading I've done over the years (which has been considerable). My problem is sort of that most things I've read suggests a lookup table for the log and anti-log, which I would agree with is the way to go when /using/ logs, but I've got a bad history with lookup tables. I can't /understand/ the thing from a lookup table, I /still/ don't understand the whole sine cosine etc thing since all I ever got was "use a lookup table". Practically yes, but I can't understand what the hell the number in the lookup table is suppose to represent if I don't understand how to calculate it.

Sorry, that was a smidgen ranty, but it's a bit of a touchy subject for me. :/

EDIT: Right after posting this I noticed this post by Kinma: https://artofmemory.com/forums/calculating-logarithms-by-hand

This post should possibly be directed there I think.

EDIT 2: Meh, copied the wrong link. -_-

Let's try to do some simple arithmetic that hopefully helps you in understanding logarithms.

it is relatively easy to calculate in base 2.

Bear with me, it will make sense in a minute.

4 times 8 equals 32.

In powers of 2:

\( 2^2 * 2^3 = 2^5 = 2^{2+3} \)

Now base 10:

100 times 1000 equals 100,000.

In powers of 10:

\( 10^2 *10^3 =10^5 = 10^{2+3} \)

I can change the base 10 in the previous line to anything random, like 7, and it will still be true:

\( 7^2 *7^3 =7^5 = 7^{2+3} \)

Let's stay with base 10.

In a multiplication \( x * y \) , if I can change both x and y to a power of 10, then the multiplication essentially becomes an addition:

\( x * y = 10^a *10^b = 10^{a+b} \)

In other words, in x * y , if I change x to 10 to the power of something and y to 10 to the power of something else, then the solution is 10 to the power of something plus something else.

Example to make this abundantly clear (I hope):

If I need to calculate 2 *3, then I can change 2 for 10 to the power of something and I change 3 for 10 to the power of something else and then I can add something to something else and then do 10 to the power of the result to find my answer.

So we need the find a and b so that:

\(2 = 10^a \) and \( 3 = 10^b \)

Turns out that \( a \approx 0.301 \) and \(b \approx 0.477 \)

And see my first couple of posts in https://artofmemory.com/forums/calculating-logarithms-by-hand to find out how to get this numbers without a calculator.

So we add 0.301 to 0.477 to get 0.778.

Then we work out \(10^{0.788} \approx 6 \).

Now the mathematical terms: 0.301 is called the logarithm of 2 and 0.477 is called the logarithm of 3 and 6 is the anti-logarithm of 0.788.

Now reread your statement:

I hope you see that my text is exactly the procedure you just described.