Based on the estimates log(2) = .03 and log(5) = .7, how do you use properties of logarithms to find approximate values for #log(0.25)#?

1 Answer
George C.
May 22, 2015

Firstly, fixing the typo in the question: #log(2) ~= 0.3#.

#log(0.25) = log(1/4) = log(1/2^2) = log(2^(-2))#

#= -2log(2) ~= -2xx0.3 = -0.6#

Incidentally, do you know why #log(2) ~= 0.3#?

#10xxlog(2) = log(2^10) = log(1024) ~= log(1000) = log(10^3) = 3#

So basically it's because #2^10 ~= 10^3#.

Related questions
See all questions in Common Logs
Impact of this question
1907 views around the world
You can reuse this answer
Creative Commons License