Speed calculators: What is really useful to memorize?

Mental Calculation, Mathematics, and Puzzles
21

Would the second log work better, faster? What would be the advantages for you?

You answers are spot on! Well done!

For log(5.5) you can use log(11) - log(2). And then you are right to ask yourself, how do you calculate log(11) when you haven’t memorized it? The answer is ‘be creative’.
Here is an idea. 11^2 = 121.
Log(120) can easily be calculated.
Log(125) can easily be calculated.
Log(11) =log(121)/2
If you calculate log(121) from a linear interpolation between 120 and 125 you will be very close.

log(120) = 2.0792 (log(3) + log(4) + 1)
log(125) = 2.0969 (log(5) - log(4) + 2)
difference = 0.0177
diff/5 = 0.00354
Add the difference: log(121) = 2.0792 + 0.00354 = 2.08274
Divide by 2: 1.04137 (log(121)/2 = log(11))

Let’s verify (using a calculator): 10^1.04137 = 10.99943
log(11) = 1.04139

It is difficult. Start with 3 digits first. You can always add more digits.

22

ok, well i just thought since it’s more whole numbers etc, as I don’t really understand how we’re going to use the memorized numbers yet, I don’t know if it will matter in the sense that we will use the base for calculating anything yet, multiples of 10 are always great for calculating.

But otherwise, I can’t see why tenth log should be better than second log? Less decimals, more doubles etc… I just want to know, if you say the tenth log is better I’ll do it, but there’s always room for improvement, and sometimes I wished I had just thought of something in the beginning and learning different.

For example log[2,5] now is just the log[2,10] - 1. I don’t know. Will it become harder for bigger numbers or create any other trouble? You tell me. :slight_smile: I have no experience in using this for mental calculations, only in real maths! I’d better decide before I big down though.

23

Base 2 is just as difficult as base 10.
You can quickly go from a base-10 log to a base-2 log by dividing the number by log(2).

So log[2,3] = 0.477 / 0.301 = 1.585.

Start with base 10. You can always switch to another base.

Now, let’s do some magic calculation.

log(2) = 0.301
log(1.1) = 0.0414 (from log(11) we just calculated)

If we divide 0.301 by 0.0414 we get 7 1/4.
This means that if we get 10% interest per year on an investment, it takes 7 years and 3 months to double your money.
Mathematically 1.1^7.25 = 2.

These are calculations that not a lot of people can do without a calculator.

24

ye definately. That’s cool. Btw I did up to 20 now.

Ok, so I get this - how to use it to solve A ** x = B type of equations (where A and B are known)
And I also figured out how to do the A ** B = x on my own. (But it seems to require some knowledge of the bigger log numbers. Maybe we should look a bit at this? I might be missing something.)

What’s next? :slight_smile:

25

What do you mean with the bigger log numbers? Can you give examples?

26

hehe why don’t you just show how to calculate something like 73 or 46, and if it’s possible: something like 6**7, but i’m guessing we will need to memorize more digits of the logs in atleast the last case, and maybe the others too?

27

If you are calculating let’s say 7.4 ^ 2.4 ~= 122 (~= is pronounced ‘is around’).

This can easily be done by taking 2.4 X log(7.4) = 2.0862.
Then we need to take the anti-log of 1.0862. This is: 10^2.0862

This is easier calculated than it seems at first sight.
Let’s split the number:
2.0862 =
2 +
0.0414 +
0.0414 +
0.0034

2 translates to 100 (10^2)
0.0414 translates to 1.1 (we calculated this, remember?)
0.0034 = 1.008.

So we have 100 X 1.1^2 X 1.008.
This is 121 plus 0.8% ~=122.

28

Here is how to do 7^6.
First we calculate the number of digits in the answer.
This tells us how precise we can get to the number and how many digits precision we need.

We do a rough calculation to see how many digits there are in the answer:
log(7) ~=0.85
6Xlog(7) = 5.1
We split 5.1 into 5 and 0.1
10^5 = 100,000
10^0.1 = 1.26
So the answer is roughly 126,000.

Now let’s do the same using 4 digits precision:
log(7) = 0.8451
6 x log(7) = 5.0706
Split the log into:
5 (=100,000) +
0.0414 (= 1.1) +
0.0292 (= 1.07, because 292/414 ~= 0.7)

100,000 X 1.1 X 1.07 = 117,700

The calculator tells us that 7^6 = 117,649
We are 51 off.

29

oh wow that’s a lot of complicated calculations. Atleast it looks like it to me. I think I could never do them in my head, maybe on paper. But it seems like it would take a while to perform even then.

30

There sometimes are different ways of calculating powers.
For example in the 7^6 calculation you could realise that 7 squared is 49 which is 50 minus 2%.
Then for 7^6 you can first calculate 50 ^ 3 = 125,000.
Then subtract the 2% 3 times.

This works with round powers.
For calculations in the form of 7.3 ^ 2.4 the way with logarithms is quickest IMO.

The calculations might seem difficult. But keep in mind that you are doing calculations that most people cannot even do.
Start easy. In the example of 7 ^ 6 above I first did an quick estimate to see in what range the answer would be.
Can you do those calculations? Surely you can multiply .85 by 6.
So your perceived difficulty probably lies in converting 5.1 into its antilog.
‘5’ means 100,000 and after that you are left with ‘.1’.

When I started out doing these calculations I started wondering if I could calculate 10 ^ .1, .2, .3, etc.
10 ^ .3 needs to be close to 2 (since log 2 = .301) so 10 ^ .1 needs to be close to the third root of 2.

Then I started thinking about the calculations from the perspective of a bank.
So I thought ‘which percentage interest, payed out 3 times, results in doubling my bank account?’.
This is the same calculation as the third root of two.

By rewording these problems you can make them easier to understand and easier to manage.

When you start out doing the estimate calculation like I showed you above, you will get a sense for these numbers.
After that, when you feel comfortable doing these calculations, you can focus on doing them with more digits.

On your smartphone, download a scientific calculator so you can check your mentally calculated answers.
This way you will learn from your mistakes.

Let’s do another. 3^10.
Log 3 = 4.77
10 X .477 = 4.77.
4 means 10,000 and we are left with 0.77
Log 6 is .778, so the answer needs to be close to 60,000.
The difference between log 6 and 0.77= 0.778-0.77= 0.008
If log 1.1 is 0.0414 then you can appreciate that log 1.01 is ten times as small or 0.00414. It is a little bigger 0.0043, but this does not matter for this calculation. This, 1.01, is 1 percent, right?

If one percent is 0.004 then 0.008 must be two percent, right?

Again, if log 6 = 0.778 and we are calculating the antilog of 0,77 and we know that 0.004 represents about one percent then we need to subtract 2 percent.

Long story short 0.77 must be around 2 percent smaller than 6.

6 minus 2 percent is 5.88
The answer must be close to 58,800.

3^10 = 59,049

See if you can follow this.
Try out some calculations and let me know where for you it becomes difficult.

31

You could try memorizing the non-trivial composite numbers below 1,000. This can be useful for factoring, and there’s only 70 of them to remember:

91, 119, 133, 161, 203, 217, 221, 247, 259, 287, 299, 301, 323, 329, 343, 371, 377, 391, 403, 413, 427, 437, 469, 481, 493, 497, 511, 527, 533, 551, 553, 559, 581, 589, 611, 623, 629, 637, 667, 679, 689, 697, 703, 707, 713, 721, 731, 749, 763, 767, 779, 791, 793, 799, 817, 833, 851, 871, 889, 893, 899, 901, 917, 923, 931, 943, 949, 959, 973, 989

Source: Tanya Khovanova's Math Blog » Blog Archive » Remember Your Primes

For an even bigger challenge, check out the 43-page Memorization chapter of the Number Sense Tricks PDF: http://parttimemusic.com/misc/NumberSense.pdf

32

kinma:

I can follow, but it seems like an awful lot of work. Now I don’t only have to remember for example 1 2 3 4, but also all of the decimals. ^^ I guess the decimal-trick can pretty much be used every time though? Although, why not use 1.2, then? lg12 = .0791…

Anyway, I guess this is a little bit lame to say, but I thought it would be less work. :slight_smile: It’s still great, just not exactly what I expected.

I’ll try one:

5**5:

lg5 = 0.6989

5 * 0.699 = 5 * 0.7 -0.005 = 3.495

3 means 1000,

that leaves 0.495. Right about now I’m starting to wish that there existed a system that better took the value of the numbers into account when creating the numbers. I can make such a system for 0-9. But all the way up to 100 seems difficult. Maybe doable… For example by making 10 different levels of some quantity, like “best movie”, and then 10 different characters from each movie. Either way, back to the calculation:

lg5 started with 0.69, so too big; try lg3: 0.4789.
Ok, that seems like it’s about as close as I will get.

10 ** (lg3.496) is a bit bigger than 10 ** (3+0.4789) = 10 ** 3 * 10 **0.4789 = 3000.

The answer should be a little bigger than 3000.

The answer is 3125. Maybe this was too inexact answer? Off by exactly 4%.

Greyhound:

Wow that books seems much better than the one I’m currently reading on speed mathematics. I should totally switch. :slight_smile:

33

Yes, the decimal-trick can always be used. lg12 can thus indeed be remembered as ‘.0791’.

The 4 % difference is the difference between 0.495 and log(3).
If you take that into account, your answer is spot on!
So it seems to me that you are on the right track to be able to mentally calculate any power that people throw at you.