How do you calculate #log 1300#?
1 Answer
May 20, 2017
Explanation:
Suppose you know:
#log 2 ~~ 0.30103#
#ln 10 ~~ 2.302585#
Then:
#1300 = 10*2*65 = 10*2*64*65/64 = 10*2^7*(1+1/64)#
So:
#log 1300 = log (10*2^7*(1+1/64))#
#color(white)(log 1300) = log 10 + 7 log 2 + log(1+1/64)#
#color(white)(log 1300) = log 10 + 7 log 2 + ln(1+1/64)/ln 10#
Now:
#ln(1+x) = x-x^2/2+x^3/3-x^4/4+...#
So:
#ln(1+1/64) = 1/64-1/(2*64^2)+1/(3*64^3)-1/(4*64^4)+...#
#color(white)(ln(1+1/64)) = 1/64-1/8192+1/786432-1/67108864+...#
#color(white)(ln(1+1/64)) ~~ 0.01550419#
and:
#log 1300 = log 10 + 7 log 2 + ln(1+1/64)/ln 10#
#color(white)(log 1300) ~~ 1 + 7*0.30103 + 0.01550419/2.302585#
#color(white)(log 1300) ~~ 3.1139434#