How to calculate antilogarithm of 0.87 through this method presented by Zach Wissner-Gross?

So, Mr. Wissner-Gross had done this video about calculating logarithms: Calculating logarithms in your head - YouTube

It’s a very good video, I understand the method BUT it’s only for calculating logarithms.

How calculate antilogarithms? What I have to do?

I appreciate the help offered by you @Kinma, although for me is difficult to catch the sense of your method ^^’

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That’s a nice video actually, very clear. it’s the same method that I have on my site here: mental calculation of logarithms.

Calculation of antilogarithms uses the same method, but backwards, and there is some guesswork required. That method is here: mental calculation of antilogarithms.

In brief, to calculate e.g. 10^0.12345, you find some combination of memorized log values that sum to 0.12345. Here the first one I see is 2 * log 2 – log 3 = 0.12494.

Then 0.12345 = 2 * log 2 – log 3 – 0.00149

Since 0.00432 represents a 1% change, 10^0.12345 = (2^2 ÷ 3) – (149/432)%

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Thanks Daniel.

However it’s only clear to this point:

0.12345 = 2 x log 2 - log 3 = 2 x 0.30103 - 0.47712 = 0.12494

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This part is hard to understand ^^’

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This part perhaps is the hardest part—it’s using an approximation to conflate addition and multiplication.

Adding 1% is the same as multiplying by 1.01.

Multiplying by 1.01 is the same as adding 0.00432 to the logarithm.

Adding 2% is the same as multiplying by 1,02. But this is approximately 1.0201 = 1.01^2.

So adding 2% is the same as adding 2 * 0.00432 to the logarithm.

In general, adding k% is the same as adding k * 0.00432 to the logarithm.

So when you have e.g. 0.00149 added (or subtracted) to the logarithm, this is less than 0.00432. In fact is is 149/432 of that. So if 0.00432% is 1% then 0.00149 is (149/432)%. In this case, a subtraction.

(2^2 ÷ 3) = 1.33333

149/432 is just over 1/3, so (149/432)% of 1.33333 is about 0.0044/ So 10^0.12345 is about 1.3289

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That is what I don’t understand…

Where the 0.00149 came from?

And why dividing by 3?

By the way, thank for your answer Daniel

Okay, first thing: adding e.g. log 7 is like multiplying by 7. So subtracting a logarithm is like dividing by that number.

In this case, we subtracted log 3, so this means we are dividing by 3.

So if we were solving 10^0.12494, we could just say “that’s 2*log 2 – log 3, so the answer is (2^2 ÷ 3) = 1.33333”.

But second thing: we want 10^0.12345, where the exponent is 0.00149 less than 0.12345. So we need to make the 149/432 adjustment described earlier.

Hope that helps with both questions!

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Yes, indeed.

Thank for your kindness!

Here is how to calculate the antilogarithm of 0.87 (the easiest way).

Recall:
log 7 = 0.845
log 8 = 3 X log 2 = 3 X 0.301 = 0.903.

So .87 can be found between 7 and 8 and probably close the the middle.

The average of the two is:
(0.845 + 0.903)/2 = 1.748/2 = 0.874

To find the middle of 7 and 8 in terms of antilogs we need to find the so called geometric mean.

7 times 8 = 56 and the geometric mean is the square root of 56.
Now 7.5 X 7.5 = 56.25.
56.25 is about half a percent too big (it is actually a little smaller than a half percent).

So the square root of 56 is about half that half percent smaller than 7.5. This is a quarter percent of course.

So a guess of 7,48 is in order.

But now we have the antilog of 0.874. And we need 0.87!
A drop of 0.004 is about 1%, as you have learned from the video, so subtract 7 cents from 7.48 and get 7.41.

So the antilog of 0.87 is 7.41.

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Thanks @Kinma!