Even easier mentally calculating logarithms

[Kinma]

In this thread I showed how to mentally calculating logarithms:

Calculating logarithms by hand

Just for fun, let’s calculate all logarithms from scratch. And let’s start with the logarithm of 2. I start with doubling 2 a couple of times: 2 4 8 16 32 64 128 256 512 1024 Now, 1024 is a number we can easily guess the logarithm from => a little over 3. In other words, 2^{10} = 1024 => log(2^{10}) = log(1024) => 10*log(2) = log(1024) = 3 + log(1.024) => log(2) = 0.3 + log(1.024)/10 So the difficult part now is log(1.024). Let’s use [Pascals triangle]: Using the triangl…

In it I wanted to show how to do this and still get accurate numbers.

If we accept a little less accuracy, can we mentally calculate logarithms quicker and easier?
The answer is yes.

Let’s start with log 3 and use 3 * 3 = 9.
We will calculate log3 by first calculating log9 starting with log10.

log10 = 1

log9. 9 is of course 10% lower than 10.

We have memorised log 1.01 as 0.0043, so log 1.1 \approx 0.043 or 10 times as much.
It needs to be slightly less, so take 0.042. (This is because 1.01^{10} > 1.1).

Using 0.042 we get:

log 9 = log (3^2) \approx 1 - 0.042 = 0.958.

The log of a squared number is double the log of that same number.

If log (3^2) \approx 0.958, then:
2 * log (3) \approx 0.958, so:
log (3) \approx 0.958 / 2 = 0.479.

We calculated 0.479, which is close to the actual value of 0.477…

Is this fun?
Let me know.

[benjamin1990 (Benjamin Sandro)]

It is but I prefer the technique I have learnt

[Kinma]

TLDR
The first post can be improved with the below realisation.
Then we calculate log 3 as 0.478, which is a pretty accurate value.

There is something I forgot.
The difference from 1 to 1.1, logarithmically, is smaller compared to the difference from 1 to 0.9.
Or +10% is bigger than -10%.

In numbers,
log 1.1 is 0.0414…
log 0.9 is -0.0458…

458 is bigger than 414. The difference is 44. Coincidence? Well, no.
0.414 -0.458 = log1.1 + log0.9 = log(1.1 * 0.9) = log(0.99)
0.99 is 1 minus 1%.

For adding 1% we used 43 (as in 0.0043).
We now see that the logarithm for subtraction is bigger in value than the one for addition, so the number for subtracting 1% should be bigger than 43, so let’s take 44.

Why 44? Well we know the number needs to be around 4343 and 1.01 * 0.99 is 0.9999, so we need to add about 1% of 4343 to get about 4386.
43.86 rounded to 2 decimals is 44.
Forget if this get’s too complicated or just ask me to clarify.
At least remember that the numbers 44 and 43 will pop up every time: log(e).

Going back to estimating log 0.9 from the first post.
If log 1.1 \approx 0.0414 then:
log 0.9 \approx -(log1.1 + log1.01) = 0.0414 + 0.0044 = 0.0458 or 0.046 when using only 2 decimals.

Long story short.
We need to use 46 instead of 42.
And then we get for log 3:
log 9 = log (3^2) \approx 1 - 0.046 = 0.956.
log (3) \approx 0.956 / 2 = 0.478.

0.478 is pretty close to the actual value of 0.477…

I started this post by saying that I forgot that log(0.9) is higher in value than log(1.1).
I stated earlier that one should not use differences bigger than 1%.
In those cases the difference is smaller and for most cases one can use log(1.01) \approx -log(0.99). Since 1.01 * 0.99 = 0.9999, the difference with 1 is very small.

[Kinma]

Now let’s do log2

Let’s calculate log2 using 2^3=8.
log10 = 1
8 is 20% less than 10.
We used 0.046 for minus 10%, so let’s use 0.092 or twice that number.

1 - 0.092 = 0.908.
log 8 = log(2^3) \approx 0.908
log8 is three times log2, so:
log2 \approx 0.908 : 3 = 0.302666... or 0.3027 rounded.

0.3027 is a bit higher than the actual value of 0.301.
Still, accurate!

(We can do better by realising that 0.9^2 = 0.81. So instead of taking log(0.8) = 2 * log(0.9) we should make a small correction 80 to 81 or 1.25%
This would give a very accurate value.)

[Kinma]

We calculated log2 on a difference of 20%. This is way too much. We know that our answer is not very accurate.

Let’s see if we can see how accurate our numbers are.
We calculated log3 to be 0.478 and log2 as 0.3027.
Then log6 = log2 + log3 = 0.7807.

Bear with me for a moment.
Let’s find log 2 in a different way:
100 / 6 = 16 + 2/3 \approx 16 * 1.04 = 2^4 * 1.04

So log(100 / 6) = 2 - log6 \approx 2 - 0.7807 = 1.2193
log(16 + 2/3) \approx log(16 * 1.04) = log(2^4) * 1.04 \approx 4 * (log(2) + log(1.01)) The last step we can do because {1.01}^4 \approx 1.04.

1.2193 = 4 * (log(2) + log(1.01))
log2 = 1.2193 / 4 - log(1.01) = 0.304825 - 0.0043 = 0.300525

So now we have 2 numbers for log 2:
0.3027 and 0.3005.

Let’s split the difference:
Log 2= 0.3016

A bit closer to the actual value of 0.301…