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Monday, August 09, 2004 - 09:33 PM
I came up with this method for determing the logarithm (base 10) of a number in my head back in 1995. I was tutoring a friend for her MCAT exams and a question involving pH and pKa values struck my interest. Although we were able to arrive at a suitable answer without actually needing a calculation, I still wanted a quick way of determining logarithms in the event that I found myself without a calculator (and just as a mental challenge for myself).Here is how I do it:
1. When someone gives you any positive number, you should immediately 'write' that number in scientific notation in your head.
2. Next, focus only on the exponent of the number (written in scientific notation). This number will be the basis of you answer.
3. Estimate the logarithm of the abscissa in your head (thats the number between 1 and 9.999999..., not part of the exponent). Note: you'll need to memorize the table below (its not that hard).
4. Add the logarithm of the abscissa to the exponent you found in step 2.
What follows are the values for the logs you'll need to have memorized for step 3...
log[1]= 0
log[2]= .30
log[3]= .48
log[4]= .60
log[5]= .70
log[6]= .78
log[7]= .85
log[8]= .90
log[9]= .95
As an example, lets find the logarithm of 29,012. Written in scientific notation, that would be 2.9012 X 10^4. So, the exponent is 4. Now, we need to concentrate on the abscissa (2.9012 is very very close to 3). From our table above (which we have memorized for the trick), the logarithm of 3 is 0.48. So, we add the exponent (4), to the log of the abscissa (0.48), to get a value of 4.48. A calculator reveals how good this method is (4.46 out to two decimal places).
This method works because scientific notation is a base 10 system for writing out numbers. With some practice, you'll get a feel for how to massage your guesses for numbers that aren't exactly in the table (for instance: 2.5).
1. When someone gives you any positive number, you should immediately 'write' that number in scientific notation in your head.
2. Next, focus only on the exponent of the number (written in scientific notation). This number will be the basis of you answer.
3. Estimate the logarithm of the abscissa in your head (thats the number between 1 and 9.999999..., not part of the exponent). Note: you'll need to memorize the table below (its not that hard).
4. Add the logarithm of the abscissa to the exponent you found in step 2.
What follows are the values for the logs you'll need to have memorized for step 3...
log[1]= 0
log[2]= .30
log[3]= .48
log[4]= .60
log[5]= .70
log[6]= .78
log[7]= .85
log[8]= .90
log[9]= .95
As an example, lets find the logarithm of 29,012. Written in scientific notation, that would be 2.9012 X 10^4. So, the exponent is 4. Now, we need to concentrate on the abscissa (2.9012 is very very close to 3). From our table above (which we have memorized for the trick), the logarithm of 3 is 0.48. So, we add the exponent (4), to the log of the abscissa (0.48), to get a value of 4.48. A calculator reveals how good this method is (4.46 out to two decimal places).
This method works because scientific notation is a base 10 system for writing out numbers. With some practice, you'll get a feel for how to massage your guesses for numbers that aren't exactly in the table (for instance: 2.5).
If you like CuriousMath.com, here's a book you'll love:
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Comments
To follow-up on your example, which you could improve the accuracy of the results without paying a heavy price with the help of interpolation.
Knowing that the logarithm of 30,000 is 4.48, and that of 20,000 is 4.30, one could easily deduce an estimate of the logarithm of 29,000 by linear interpolation, namely subtract one tenth of the difference between 4.30 and 4.48, or
Log(29000)~= 4.48 - (4.48-4.30)/10
= 4.48-.018=4.462
Correct value for log(29000)=4.46240
Correct value for log(29012)=4.46258
==============================
For those who have extraordinary memory, you could memorize the logarithms to four digits as follows:
log(1)=0
log(2)=.3010
log(3)=.4771
log(4)=.6020
log(5)=.6990
log(6)=.7781
log(7)=.8451
log(8)=.9031
log(9)=.9542
Happy calculations!
So in base2, log2(29,012):
(log10(2.9012)+4) / log10(2) = 14.824....
log[2] = 0.3
log[3] = 0.48
log[4] = log[2] + log[2]
log[5] = 0.70
log[6] = log[2] + log[3]
log[7] = 0.85
log[8] = log[2] + log[2] + log[3]
log[9] = log[3] + log[3]
Memorizing only the primes helps even more when you are memorizing 2 digits, such as the log of 1.7. There are only 23 numbers to memorize then!
This is probably a typo -- it should be
log[8] = log[2] + log[2] + log[2]
since 2*2*2 = 8.
Remember that 10/5 = 2, so you only need remember the logs of 2, 3 and 7:
log[2] = 0.30103
log[3] = 0.47712
log[4] = log[2] + log[2]
log[5] = 1 - log[2]
log[6] = log[2] + log[3]
log[7] = 0.84510
log[8] = log[2] + log[2] + log[3]
log[9] = log[3] + log[3]
Some people train their memories. For instance, a digit can be represented as letters and then the letters form words or acrostics. So the building block logarithms can be learned to many digits without a lot of strain. You can read more about turning letters into digits at www.NakedScience.com
Mike
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