Logarit2.rtf - Derivation of a log using Euler's method. November 20,21, 2019, afk
Note: this paper is a fleshing-out of the following link:
http://lamarotte2.blogspot.com/2011/04/calculating-log-of-number-by-hand.html
Problem: Derive log(base 10) n
Method:
1.)Get highest power of 10 which is below n : (low)
2.)Multiply this value by the lowest power of 10 which is above n: (low)(high)=p
3.)Extract the square root of this product. √p
4.)Derive the logarithm of √p ; ( in two steps.)
( log low + log high ) =log p ; First : logarithm of the entire product : log10 p.
log p /2 =log √p ; Then : logarithm of the square root of the product : log10 √p
5.)Repeat Method using the values which are closest in value to n until the value of the converging results
is close enough to the desired precision and accuracy.
Note : Simple powers of 10 have easily-calculated logarithms 10/1, 100/2, 1000/3, etc. By selecting powers of 10 immediately below
and above n we set a high and low boundary each of whose logarithms can be established exactly. Each recurse through
the algorithm resets the previous boundary to a new boundary which is closer to n.
The method concludes when √p =n to the desired precision and accuracy.
Example: Derive log (base 10) 749
1.)Get highest power of 10 which is below n
The highest power of 10 which is below 749 is 100 (low)
2.)Multiply this value by the lowest power of 10 which is above n (high)
The lowest power of 10 which is above 749 is 1000
100 x 1000 = 100,000 =p
3.)Extract the √p
√100,000 = 316.227
4.)Extract log √p ( (log low + log high) / 2 ).
log 100 =2
log 1000 =3
2 + 3 = 5 (Log p =5)
Divide log p by 2 to get log √p
log √p =5/2 =2.5
Summary and check:
p =105 =100,000
log p =5 (105 =100,000)
√p =316.227
log √p = log p /2 =(5/2) =2.5
102.5 =316.227 =√100,000 =√p
;------------------------------------------------------------------------------------------
5.)Repeat Method using the values which are closest in value to n until the value of the converging results is close enough to the desired precision.
The derived value 316.277 is below 749 but above 100 so it becomes the new low value
to be plugged into the Method. The original high (1,000) remains the only value greater
than 749 but <= 1,000.
On Entry :
low =316.277
high =1,000
(2a.) (low)(high) =(316.277)(1,000) =316,277 =p
(3a.)Extract √p
√p =√316,277 =562.385
(4a.) log low +log high =log p
log low =log 316.287 =2.5
log high =log 1000 =3
log p =2.5 +3 = 5.5
log √p =5.5 /2 =2.75
;------------------------------------------------------------------------------------------
(Repeat Method) The new square root of 562.385 still isn't 749 so we continue. Because
562.385 is lower than 749 it now becomes our new low
On Entry : low =562.385
high =1,000
(2b.) (low)(high) =(562.385)(1,000) =562,385 =p
(3b.)Extract √p
√p =√562,385 =749.923 (This is close enough for this tutorial)
(4b.) log low +log high =log p
log low = log 562.385 =2.75
log high =log1000 =3
log p =2.75 +3 = 5.75
log √p =5.75 /2 =2.875
;----------------------------------------------------------------------------------------------
102.875
A TI-34 calculator shows log 749 =2.874481818 and 102.875 = 749.8942093
Accepting 2.875 means that we rounded the last 7 digits (2.874481818) up to 2.875 which
is acceptable precision for this tutorial. If greater precision is needed then repeat the process.