Is there an easy way to calculate antilogarithms?

Mental Calculation, Mathematics, and Puzzles
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Is there an easy way to calculate antilogarithms?

References:

[note: some posts below were removed]

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Paging @kimna lover of logs

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Do you want to know how antilog work, the use and such?
You already found the link on how to calculate this mentally, so let’s give a short primer on logs and antilogs:

if I can write a number x as a power of - let’s say - ten, we can say:

x = 10^a

Add a second number y:

y = 10^b

If we multiply these 2 numbers a curious thing happens:

x*y = 10^a * 10^ b = 10^{a+b}

We start with a multiplication and end in addition!

Example:
100 * 1000 = 10^2 * 10^ 3 = 10^5 = 100,000

a is called the logarithm of x.

And

x is called the antilogarithm of a.

An example:
10^2 = 100
So the logarithm of 100 is 2.
Reversely; the antilogarithm of 2 is 100.

10^{0.3} \approx 2
So the logarithm of 2 is approximately 0.3.
Reversely; the antilogarithm of 0.3 is approximately 2.

This is true in base 10, meaning that we use the number 10 to raise numbers with.

This is also true in other bases.
The logarithm (base 2) of 8 = 3, because 2^3 = 8
The antilogarithm (base 2) of 3 = 8.

Base ‘e’ is often used. Where e is Eulers number, roughly 2.71828183. This is called the natural logarithm. Often the ‘ln’ key on the calculator.

Hope this helps. If you have further questions or want to know more about how to mentally calculate these things, just ask.

In the link I talk about mentally calculating antilog base 10. We can do e too.
So if you studied the link you posted and want to become a natural (anti)logarithm wizard, I will make a similar post how to do this.
But you have to understand the link you posted first.

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Your answer tells me that you don’t actually understand antilogs.
This is ok, it is complicated stuff.

However; the answer to this is easy.
It is a one with 37 zeros or:

10,000,000,000,000,000,000,000,000,000,000,000

I explain how to mentally calculate the first 21numbers here:

So for 37, just extrapolate.

However; again I think you want to know something else.

The antilog of a number is the base number to the power of the number.
So assuming base 10, the antilog of 37 is 10^{37}.

You now have the links how to calculate both log and antilogs.
You can apply this to calculate log 37 .

Hint: since 37 is prime, I would take the middle of log 36 and log 38.
Then use 36 = 6*6 and 38 = 19*2.

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Yes, the 2% is just the difference between 50 and 49.

My reasoning was:
It is difficult to directly calculate log 7.
However; the logarithm of 7^2 = 49 is easier, since it is close to 50, which is 100/2.
And for 100 and for 2 we know the logarithms.

So if one can calculate log50 and make a change to move to log 49, then one has a very accurate calculation (assuming one has an accurate value for log2.

We can do the same with 37:
37^3 = 50,653 which is just over 1% higher than 50,000. (1.3% if you want to be precise).

So find log 50,653 and divide it by 3.
50,000 = 100,000/2 so the log of 50,000 =
5 - 0.301 = 4.699

Add 1.3%
1.3 * 0.0043 = 0.00559
4.699 + 0.00559 = 4.705 (rounded to 3 decimals)
4.705/3 = 1.568

Here is how I divide 4.705 by 3:
1.5 times 3 equals 4.5, so subtract that from 4.705 to get 0.205. Now in my mind I multiply 0.205 by one thousand to get 205. 210 is 70 times 3. Subtract 3 twice from 210 to get 204. And I hope you immediately see that that is 68 times 3.

So now we have 1.568 with a remainder of 1.
Either call out 1.5683333… or round to 3 decimals to get 1.568.

A calculator gives: log 37 = 1.568

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Sure!

While thinking about 37^3 I came across an interesting way of doing this, making the calculation easier (I hope).

First, let’s square 37 using (40-3)^2:
37 * 37 = 34 * 40 + 9 = 1,369

Now 1,369 is almost 1,370.
Let’s do 3 times 1,370 first.
Forget about the last zero for now and do 3 x 137.
3 x 13 = 39 and 3 x7 = 21, so the multiplication gives:
39 | 21
Take care of the carry to get:
411

If 3 x 137 = 411, then 3 x 1370 = 4,110. Done.

(as always, a lot of work to write out; goes quick in your mind)

Alternatively do:
3 x 1,300 = 3,900
3 x 70 = 210.
add 210 to 3,900 to get 4,110.

Now this is 3 X1,370. We need 3 X 1,369, so subtract 3 to get 4,107.

Next step: 7 X 1,369.
We could multiply 7 times 1,369, but what if I told you a quick subtraction is enough?

Because 7 =10 -3, we can do:
7 X 1,360 = (10-3) * 1369 =
13,690 - 3 X 1,369 =
13,690 - 4,107 (Remember we just calculated 3 * 1,369 = 4,107, so no need to do that again)

In your mind, just add a zero to 1369 and subtract the number you just calculated doing 3 times 1369 (4,107).

13,690
4,107 -
_ _
9,583

If 3 times 1369 = 4107, then 30 times 1369 = 41,070
Now add 7 times 1369 = 9,583

41,070
9,583 +
_ _
50,653

It takes some time to write this out and maybe it takes some time for you to do this mentally,
but how many of your friends can do 37^3 mentally?

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Sure. Understandable.

In general, taking the middle of the next adjacent numbers almost always gives satisfactory results.

Example:
11^2 = 121
10*12 = 120.
One off.

So if you want to calculate log11, take log120 first.
log 120 = log3 + log4 + log10 =
0.477 + 0.602 + 1 =
2.079

Make the correction of 1 in 121 or about 0.8%:
0.8 * 0.0043 = 0.0032 + 0.00024 = 0.00344

2.079 + 0.00344 = 2.082 (rounded)
so:
log121 = log(11^2) = 2 * log11 \approx 2.082

Divide by 2:
log 11 \approx 2.082 / 2 = 1.041

A calculator gives log 11 = 1.0414.

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That is an easy one.
log 2 = 0.301
So 10^{0.315} = 2 * 10^{0.014}. (subtracting 0.301 from 0.315).
If 1% equals 0.0043 in log terms, then 0.014 is about 3 times as much. 3.25 if you want to be precise.

So add 3.25% to 2 and get:
2.065

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In general, look for one of the logs you have memorized. Then look at the difference and determine a percentual difference.

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