Is there an easy way to calculate antilogarithms?

[Kinma]

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

Take, for example, the antilogarithm of 37.

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

First, I should know how calculate the logarithm of 37.

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

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…

So for 37, just extrapolate.

Unfortunately, all the sources I found explains antilogarithms in a way I can`t understand

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.