How do you calculate mentally an antilogarithm, for instance antilogarithm of 10^0.93?

antonio1990k

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Hi,

I would like to know how calculate in my head antilogarithms.

I'm very bad at reasoning so if you can teach me about the method I will be grateful.

Thanks
 

HallsofIvy

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Except for special cases, there is NO way of calculating antilogarithms in your head! In fact, there is no simple way to calculate antilogariths with paper and pencil. If \(\displaystyle y= log_a(x)\) then the antilogarithm is \(\displaystyle x= a^y\). Except when x is an east power of a, that will be difficult to calculate. For example, if a= 10 (the "common" logarithm) then the antilogarithm of 2 is \(\displaystyle 10^2=100\), of 3 is \(\displaystyle 10^3= 1000\), of -1 is \(\displaystyle 10^{-1}= 1/10\), etc.. If the base is "e" (the "natural" logarithm) then the antilogarithm of -1 is \(\displaystyle e^{-1}\), etc.

When I learned logarithm, about 10 B.C. ("Before Calculators") we looked up logarithms and anti-logarithms in tablles (who here remembers the "C.R.C" ("Chemical Rubber Company") tables?)
 
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antonio1990k

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Except for special cases, there is NO way of calculating antilogarithms in your head! In fact, there is no simple way to calculate antilogariths with paper and pencil. If \(\displaystyle y= log_a(x)\) then the antilogarithm is \(\displaystyle x= a^y\). Except when x is an east power of a, that will be difficult to calculate. For example, if a= 10 (the "common" logarithm) then the antilogarithm of 2 is \(\displaystyle 10^2=100\), of 3 is \(\displaystyle 10^3= 1000\), of -1 is \(\displaystyle 10^{-1}= 1/10\), etc.. If the base is "e" (the "natural" logarithm) then the antilogarithm of -1 is \(\displaystyle e^{-1}\), etc.

When I learned logarithm, about 10 B.C. ("Before Calculators") we looked up logarithms and anti-logarithms in tablles (who here remembers the "C.R.C" ("Chemical Rubber Company") tables?)
Hi,

There's a video on YouTube about mentally calculate logarithm:

I guess that in order to calculate antilogarithm we should operate backwards but I can't...
 

Dr.Peterson

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I would like to know how calculate in my head antilogarithms.

I'm very bad at reasoning so if you can teach me about the method I will be grateful.
I presume you meant, to calculate the antilogarithm of 0.93, which is 10^0.93.

Do you have some reason to think this can be done in your head, in general? I know there are parts of the world where calculator usage is not expected, and tables are still used; but I don't know of any non-table methods. In fact, the first publications of the concept of logarithms included tables, because they just can't be done directly (except, as mentioned, for integers).

So the best I can say is, write down a 1 followed by 0.93 zeros. It will look something like "1C". ;)
 

antonio1990k

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I presume you meant, to calculate the antilogarithm of 0.93, which is 10^0.93.

Do you have some reason to think this can be done in your head, in general? I know there are parts of the world where calculator usage is not expected, and tables are still used; but I don't know of any non-table methods. In fact, the first publications of the concept of logarithms included tables, because they just can't be done directly (except, as mentioned, for integers).

So the best I can say is, write down a 1 followed by 0.93 zeros. It will look something like "1C". ;)
It can be done mentally, but I don't understand the steps: https://worldmentalcalculation.com/how-to-calculate-antilogarithms/
 

HallsofIvy

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Did you understand the very first statement at that site:

"Using the memorized values for various logarithms"?

In other words, you first memorize a lot of values, then use things like
"log(x)+ log(y)= log(xy)"
"log(x)- log(y)= log(x/y)" and
"y log(x)= log(x^y)"
to find other antilogarithms/
 

antonio1990k

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Did you understand the very first statement at that site:

"Using the memorized values for various logarithms"?

In other words, you first memorize a lot of values, then use things like
"log(x)+ log(y)= log(xy)"
"log(x)- log(y)= log(x/y)" and
"y log(x)= log(x^y)"
to find other antilogarithms/
Yes, I understand I must memorize some logarithmic values.

Nevertheless, I'm unable to approximate the chosen number to the logarithm and then correct the approximation :/
 

JeffM

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The whole thing is stupid.

“Using the memorized values” already tells us that thinking is not required. Like Halls, I still have my CRC book of tables, which eliminated the need for memorizing. Nowadays, much of what was in those tables is in my calculator. Mental computation of log10(82} has no practical utility at all.

I do not deny that mental computation of basic arithmetic operations has some utility, but there is no practical utility or mathematical insight in computing MENTALLY an approximation of antilog10}(2.812).
 

Dr.Peterson

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Nevertheless, I'm unable to approximate the chosen number to the logarithm and then correct the approximation :/
Which step of which example do you not follow? Maybe we can help with that. We're not going to recommend learning this "method" for practical purposes, but maybe it's good mental exercise for you.

“Using the memorized values” already tells us that thinking is not required.
Actually, I think the problem is that more thinking is required than a rote technique (like addition) is expected to need. You have to choose carefully among the many numbers you've memorized, and then use the known properties to relate them, not in a purely rote way (as shown by the difference in the examples). So maybe it provides practice in thinking about logs.
 

antonio1990k

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The things I know is that for every 1% we add or we subtract, we add or we subtract 0.00432%, right?

In this file about mental calculation techniques, we have at page 28 the technique for calculating antilogarithms: http://www.myreckonings.com/wordpress/wp-content/uploads/LightningCalculationCalendar2011/2011LightningCalculationCalendar.pdf

Of course I still don't understand the steps but I think they are easiest than the previous post I wrote.

At step 1 the author say: Extract powers of 10 from N (by subtracting or adding integers) to leave the smallest difference from zero

10^0.93 will be 10^1/10^-0.07 then if Im'm right?

I know it's a difficult problem, and that's why I'm asking for the help of the Forum.
 

Dr.Peterson

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The things I know is that for every 1% we add or we subtract, we add or we subtract 0.00432%, right?
Not quite. What it says is, the correction factor (multiplier) is "approximately 1% for every 0.00432 in the error". The latter number is not a percent.

Can you see where they got that number from? Since the method requires understanding, you need to be able to derive it. (In fact, that's the main reason I'm pursuing this.)

At step 1 the author say: Extract powers of 10 from N (by subtracting or adding integers) to leave the smallest difference from zero

10^0.93 will be 10^1/10^-0.07 then if I'm right?
Not quite, but you probably mean the right thing. It's 10^1/10^0.07, or equivalently 10^1*10^-0.07. This is the first step in both methods, and is the most obvious piece. (In fact, it's the reason for using common logarithms in the first place.)
 

antonio1990k

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Not quite. What it says is, the correction factor (multiplier) is "approximately 1% for every 0.00432 in the error". The latter number is not a percent.

Can you see where they got that number from? Since the method requires understanding, you need to be able to derive it. (In fact, that's the main reason I'm pursuing this.)


Not quite, but you probably mean the right thing. It's 10^1/10^0.07, or equivalently 10^1*10^-0.07. This is the first step in both methods, and is the most obvious piece. (In fact, it's the reason for using common logarithms in the first place.)
It come from the YouTube video I posted, minute 2.01, the author said:

"For every 1% a number increases, its logarithm increases by log of 1.01 = 0.00432"
 

antonio1990k

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Not quite, but you probably mean the right thing. It's 10^1/10^0.07, or equivalently 10^1*10^-0.07. This is the first step in both methods, and is the most obvious piece.
Ok, right.

Next step, I chose the right one, would be:

Step 2: Add or subtract up to two copies of log3 =0.47712 to make thenumber close to a multiple of0.1. You can determine the right number of copies by looking at the second and third digits of the number, where adding log3 is like subtracting 23 in those digits, and vice versa.

That's incredible difficult for me to understand...
 

Dr.Peterson

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Next step, I chose the right one, would be:

Step 2: Add or subtract up to two copies of log3 =0.47712 to make thenumber close to a multiple of0.1. You can determine the right number of copies by looking at the second and third digits of the number, where adding log3 is like subtracting 23 in those digits, and vice versa.

That's incredible difficult for me to understand...
I think "the right one" means "the method on the right", labeled the "MacIntosh-Doerfler method", not "the correct way"!

The way to understand something like this, unless it is very poorly written, is to keep reading. Right after where you stopped quoting, they give an example that illustrates what they mean. I also suspect that some of the ways of thinking used here may have been taught earlier (I have no patience to try to read all of it!). The idea, as I see it, is that they want to add a multiple of 0.47712 to -0.14749 to get something looking like x.x0, a near-multiple of 0.1. We are starting with x.x47, and adding x.x77 to it, which has the same effect in those digits as subtracting x.x23 (because 0.077 = 1-0.923). Since 2*23 = 46, and we want 47, subtracting 2*0.47712 will come close. And in fact, we get -0.14749+2(0.47712) = 0.80675, which is close to 0.80, a multiple of 0.1.

Since this is your project to learn these "methods", just keep at it. If you find this source too hard, find another, or set a different goal. I personally see no value in it, but if it exercises your mind in a way that you find beneficial (and doesn't hurt too much), go for it.
 

antonio1990k

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I think "the right one" means "the method on the right", labeled the "MacIntosh-Doerfler method", not "the correct way"!
Yes, sorry, I mean the method on the right, not the correct way; my English sometimes is bad, as is not my maternal language.
 

antonio1990k

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I personally see no value in it, but if it exercises your mind in a way that you find beneficial (and doesn't hurt too much), go for it.
I'm interested in mental calculation, and it's in my opinion a good way to train the mind.
 

antonio1990k

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The way to understand something like this, unless it is very poorly written, is to keep reading. Right after where you stopped quoting, they give an example that illustrates what they mean. I also suspect that some of the ways of thinking used here may have been taught earlier (I have no patience to try to read all of it!). The idea, as I see it, is that they want to add a multiple of 0.47712 to -0.14749 to get something looking like x.x0, a near-multiple of 0.1. We are starting with x.x47, and adding x.x77 to it, which has the same effect in those digits as subtracting x.x23 (because 0.077 = 1-0.923). Since 2*23 = 46, and we want 47, subtracting 2*0.47712 will come close. And in fact, we get -0.14749+2(0.47712) = 0.80675, which is close to 0.80, a multiple of 0.1.
If we add 2 times 0.47712 we get 0.95424, and adding the 0.07 we get 1.02424, which is close to 1.00, a multiple of 0.1.

I'm right? :unsure:
 

Dr.Peterson

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If we add 2 times 0.47712 we get 0.95424, and adding the 0.07 we get 1.02424, which is close to 1.00, a multiple of 0.1.

I'm right? :unsure:
I don't know whether that's the best choice (and I'm not currently working on that with you, so as to know whether you're making the same choices I would). I would not expect that using 2 log(3) would be the best choice always, but maybe it is in this case.

Actually, a quick looks suggests that you use 3 log(3) instead.

Keep going and see if the answer you get (using either of these choices) is a reasonable approximation to the exact answer. I imagine part of any approximation scheme like this is learning what's good enough!
 

antonio1990k

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Ok, thank you for your patience and kindness, I will try myself until I get some correct answer.
 
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