Mentally calculating logarithms


#1

In the thread about calculating higher powers, we talked about the importance of logarithms.

In my view this is one of the key skills in mental calculation.
It makes a lot of difficult calculations easier.
For example:
Taking the square root of a number becomes division by 2.
Multiplication becomes addition.
Any power of a number can be found by multiplication.
Etc., etc.

Being able to mentally manipulate numbers is the basic skill.
To be good at mental calculation one needs to have the ABC-mentality: Always Be Calculating.

In the other thread I explained how to calculate powers by only remembering 6 logarithms: 2, 3, 7, 1.1, 1.01 and 0.9).
Actually, the numbers 2, 3, and 7 also do not need to be memorized, since one can always calculate them.
For speed, the more numbers are memorized the better.

If more people are interested in this, we could exchange tips on doing certain calculations.


#2

Here’s a good intro for those already not familiar with logarithms:

…and here’s a couple of links on doing mental logarithms:
http://curiousmath.com/index.php?name=News&file=article&sid=32
http://curiousmath.com/index.php?name=News&file=article&sid=43


#3

Ok, let’s talk about mentally calculating logarithms.
Let’s see if we can derive log 2.

Recall from this thread that log 1.01 = 0.0043. This is the log for ‘plus 1 %’.
Let’s assume we only memorize this number.

Let’s look at the evolution of 2^n:
1:2
2:4
3:8
4:16
5:32
6:64
7:128
8:256
9:512
10:1024

So 2^10 = 1024. Now 1024 is a number that makes it easy to calculate the log for.
Let’s see if we can guess the log 1024.
Log 1000 = 3, so it must a little bit more than 3.
1024 = 2.4% more than 1000.
If log 1% = 0.0043 then 2.4 must be close to 2.4 times .0043.
2.4 X .0043 = .01032.
So we can estimate that log 1024 = 3.01032.
Recall that 1024 = 2^10.
log (2^10) = 10 times log 2.
If log 2^10 = 3.01032, then it follows that log 2 = 3.01032/10 = 0.301032.

If we look at the exact number, we see that log 2 = 0.301030.
See how close we can get?


#4

Here’s a great video working out base 10 logs, including the 1% and 10% adjustments:


#5

Excellent find!
It seems the guy in the video uses the same technique as I do.

Is it just us interested in this?
I can explain how I calculate a lot of numbers mentally, but it seems not a lot of people are interested.

At some point in my life (nurturing my nerd side) I wanted to know if I could mentally calculate all numbers from 1-100 using only log(2) as a starting point.
Now I can calcutate all numbers using no memorized numbers.


#6

Earlier in this thread we calculated the logarithm of 2, just by multiplication by 2.

Can we calculate log(3) doing the same? The answer is yes.

Let’s look at the evolution of 2^n, only this time we take it a bit further:
1:2
2:4
3:8
4:16
5:32
6:64
7:128
8:256
9:512
10:1024
11: 2048
12: 4096
13: 8192
14: 16384
15: 32768
16: 65536

Now look at the evolution of 3^n:
1: 3
2: 9
3: 27
4: 81
5: 243
6: 729
7: 2187
8: 6561

This is an interesting situation: 65536 is almost 10 times as big as 6561!
Let’s assume that it is exactly 10 times bigger and see what value of log(3) this gives.

2^16 = 10 X 3^8 starting point
log(2^16) = log(10 X 3^8) take the logarithm of left and right
16 X log(2) = log(10) + 8 X log(3) take the exponents out
16 X log(2) = 1 + 8 X log(3) evaluate log(10)
2 X log(2) = 1/8 + log(3) divide by 8
log(3) = 2 X log(2) -1 /8 subtract 1/8 and swap sides
log(3) = 2 X 0.30103 - 0.125 evaluate log 2
log(3) = 0.60203 - 0.125 do the multiplication and sutract
log(3) = 0.47703 done.

Using a calculator we find log(3) = 0.47712
The difference is 9 in 100,000.
In my book that is pretty close.


#7

Hello Kinma, I’m new to the forum and mental calculation is (another) interest of mine that I would like to foster! I’m in the process of memorizing the squares of 2 digit numbers - just for fun :p. Although I think I’m particularly fascinated by mental division as to those on the outside it just seems unfathomable - I especially like cyclic numbers because you can practise mental calculation and memory!

Don’t know much about logarithms but would like to learn so to answer your question: it is not just you two interested!


#8

Welcome, CRM!

As far as 2-digit squares go, you don’t have to memorize them. There are surprisingly simple ways to quickly calculate them. Here’s my tutorial that covers both memory and mathematical approaches:

I also put together a mental division tutorial…

…but this video shows how to handle decimals up to 15:

As for starting with logarithms, get the basics down first. Here’s a good starting way to think of them…

…and here are some good intuitive and conceptual way to think of exponents and logarithms:
http://betterexplained.com/articles/understanding-exponents-why-does-00-1/
http://betterexplained.com/articles/using-logs-in-the-real-world/

Those last 3 links don’t cover much in the way of mental math with logarithms, but you want to understand what they are and how to use them before you deal with the short cuts!


#9

It is an excellent tool to have.
In meetings I can estimate a number long before anybody else can calculate it.
I am not faster than a calculator though, I just love to guess a number with a little bit of accuracy.

Logaritmhs are fascinating!
They turn multiplication into addition, division into subtraction, squaring into multiplication (by 2) and square roots into division (by 2).
In other words, they turn what’s difficult into what’s easy.

And for that reason they should be a staple for the mental calculator.


#10

Thanks for the links greymatters. The first two links were actually what got me interested in mental calculations a while ago! The video with the calculations to 14/15 was cool also. Not read all of the logarithm pages yet - need a cup of tea to get my head round these I think!

Kinma, I know what you mean! I’m just getting started in mental calculation and I’m already impressed with myself :slight_smile: I read the post about calculating higher powers with logs and the accuracy wasn’t 100%. What about calculating a single 4 X 4 multiplication - what would the accuracy be like for that?


#11

A 4X4 calculation has 8 digits accuracy.
For mental calculation most people store or calculate 5 digits at most.
So expected accuracy is about 5 digits.


#12

Hi Kinma

Do you know what kind of calculation Can I do if I know the logs(10) Up to 1000 with 4 decimals of precision?


#13

Hi Jorge,

Calculations with logs typically involve growth.
Examples are:

  • How much money do I have if I invest $1,000 into a fund that pays 5.4% in dividend, I reinvest the dividends and I do this for 10.5 years?
  • If I continue doing this, when do I double my money?
  • When do I triple my money?

I am not sure if this answers your question.
Let me know if not.


#14

Yes, it answers my question, but also, i would like to know how are these calculation aplied?


#15

I know how to use them for multiplication and other stuff, but not in problems like the one that you mentioned above.


#16

As an example, let’s do these calculations.
Let me know if any of this is unclear.

log 1.1 = 0.0414
log 1.01 = .004321

Your original assumption was to have all numbers 1-1000 memorized.
If this is the case, you will find these two numbers at 11 and 101:
log(1.1) = log(11) -1.
and log(1.01) = log(101)-2.

The calculation is 1.054^10.5.
Take the log of this:
log(1.054^10.5) =
10.5 X log(1.054).

How do we calculate log(1.054)?
If you have 1-1000 memorized, one option is to interpolate between log(105) and log(106):
log(105) = 2.0212
log(106) = 2.0253

The difference between the 2 numbers is 0.0041.

log(105.4) = log(105) + .4 X 0.0041 = 2.0212 + 0.00164 = 2.02284.
log(1.054) = log(105.4) - 2 = 0.02284.

Another option is to take log(1.1) and divide this by 2 to get to 5%, then add 8% to move from ‘5’ to ‘5.4’.

Now multiply by 10.5:
0.02284 * 10.5 = 0.2398

At 174 in your list of memorized numbers you find:
log(174) = 2.2405, then
log(1.74) = 0.2405, so

10^ 0.2398 = a little less than 1.74.

How much less?
Well 0.2405 - 0.2398 = 0.0007.
0.0007 / 0.004321 = 1/6 (actually, 7/42 = 1/6, but this is close enough)
if 0.004321 stands for 1% then 0.0007 stand for 1/6%.
1.74 - 1/6% = 1.7371
The calculator gives: 1,73710637539

$1,000 X 1.7371 = $1,737.10

Your $1,000 grows till $1,737.10

The ‘rule of 72’ is based on the fact that 72 times 1% compounded gives 100% or twice your money.

The dividend was 5.4% = 5.4 times 1% we can do an easy divide of 72 / 5.4 = 13 1/3 or 13 years and 4 months.

Use the rule of 114.

114 / 5.4 = 21 1/9 or a little over 21 years and one month.

The last 2 calculations did not use logarithms directly.
They are derived from logarithms however.

rule of 72
log 2 = 0.301
log 1.1 = 0.0414
0.301 / 0.0414 = 7.2
7.2 X 10 = 72.

Rule of 114:
log 3 = 0.477
log 1.1 = 0.0414
0.477 / 0.0414 = 11.4
11.4 X 10 = 114.
(actually, the calculation gives 115, but I wanted to be close to the Wikipedia page. :slight_smile: )


#17

Thanks, now one more question: the method I’m using to memorize the logs is by memorizing the difference of a Log starting from a base number, for example: (the numbers used in"baseN+#" Are the fourth decimal place)

Log (990): 2.9956 (base number)
Log (991): 2.9960 (base number+4)
Log (992): 2.9965 (base number+9)
Log (993): 2.9969 (base number+13)

And so on…

What do you think about it?


#18

This works really well.
I can explain to you how to calculate the differences.
That way you do not have to memorise them, only the base numbers.

If you look at your numbers the differences are about 4.3 apart: 4, 9, 13.
The factor from 990 to 991 is 991/990. This is close to 1.001.
The numbers 1.1, 1.01, 1.001 are easy to remember.
I’ll write them down. See the pattern?

Log 1.1 = 0.0414
Log 1.01 = 0.004321
Log 1.001 = 0.0004341
Log 1.0001= 0.000043427
Log 1.00001 = 0.0000043429

I hope you see where the 4.3 difference comes from.
Now, how to quickly calculate this difference?
Estimate the factor of the change from base to the number you need.
Multiply this number by 4.3. Done this is the difference.

Example.
Log 250 = 2.3979

To calculate log 251 we estimate the factor:
251/250 = 1.004
We calculate 4 times 4.3 = 17.
Add 0.0017 to 2.3979 to get 2.3996.

Let me know if this makes your system easier.

If you want more background I can also explain where this 4.3 number comes from.
This helps in memorizing this fantastic number.


#19

This is better than memorizing hundreds of numbers. It makes the system easier.

about the 4.3, i think it comes from the result of Log 1.01: 0.00"43"21 (I’m not sure)
also, Is there an easy way to find the factor?


#20

The 4.3 comes from log(e) = 0,4342944819.

Most people don’t know where ‘e’ comes from. e comes from a series of interest payments, like this:
If I put one dollar in the bank and I get 100% interest, after one year I have 2 dollars.
I can ask myself how much money I have if I get interest payed out twice a year instead of once a year.
Intuitively a lot of people think it does not matter wheter we get twice 50% or once 100%.
But they forget the compounding effect of interest on interest.

Let’s do the calculation. If I get 100% per year I get 50% per half year.
So after half a year I get 50% on 1 dollar = 1 dollar 50 cents and after the second half year I get 50% on the 1

dollar 50 = 75 cent, making a total of 2.25.
The calculation is:
(1+1/2)^2

If I do this for 4 quarters the calculation becomes:
(1+1/4)^4

You might see a pattern here:
If I get paid n times, the calculation becomes:
(1+1/n)^n

If n becomes bigger and bigger, the result for this calculation tends to go to ‘e’:

1 2.0000
2 2.2500
4 2.4414
10 2.5937
50 2.6916
100 2.7048
1,000 2.7169
10,000 2.7181
100,000 2.7183
1,000,000 2.7183

The log of this number then also tends to go to ‘43’:
1 2.0000 0.3010
2 2.2500 0.3522
4 2.4414 0.3876
10 2.5937 0.4139
50 2.6916 0.4300
100 2.7048 0.4321
1,000 2.7169 0.4341
10,000 2.7181 0.4343
100,000 2.7183 0.4343
1,000,000 2.7183 0.4343

Around 50 the number is 0.43.
It means that for differences around 2% we can take 0.43.
For the smallest of differences the number is just 1% bigger, 0.4343.
For big differences like 10%, the number is slightly smaller, 0.41…

For practical purposes I usually take 42.
42 is divisible by 2, 3, 6, and 7 and this makes it very handy for mental calculation.

An example will make this clear:
Let’s say you memorized log(60) and want to calculate log(61):

61 = 60 X (1+1/60) therefore
log(61) = log(60) + log(1+1/60)

Using ‘42’:
log(1+1/60) ~= 42/60 ~= 7/10 ~= .7

Using ‘43’:
log(1+1/60) ~= 43/60 ~= 7/10 + 1/60 ~= .717

It is a long story to explain.
However, mentally it goes like this:
61 is 1/60 more than 60. 42/60 = 7 (forgetting about the decimal point).
Now set the decimal point:
1/60 is a little over 1%, so my answer should be a little lower than 1%.
‘7’ should be 0.007.

Then I realise that for numbers smaller than 2% I should take 43 instead of 42.
So I add 1/60 to 7 to get to ‘717’ or 0.00717 with the decimal point.