How do you calculate #log 1300#?

1 Answer
George C.
May 20, 2017

#log 1300 ~~ 3.1139434#

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#

Related questions
See all questions in Common Logs
Impact of this question
6379 views around the world
You can reuse this answer
Creative Commons License