In this thread I showed how to mentally calculating logarithms:

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.