How do you calculate #log_2 (3.16)#?

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Answer 1 · 2016-05-05T22:32:38

#log_2(3.16) = log(3.16)/log(2) ~~ 1.66#

Using the change of base formula, we have:

#log_2(3.16) = log(3.16)/log(2)#

One way of finding an approximation for this is to note that:

#sqrt(10) ~~ 3.16227766 ~~ 3.16#

So:

#log(3.16) ~~ log(10^(1/2)) = 1/2#

Also use:

#log(2) ~~ 0.30103#

Then:

#log_2(3.16) = log(3.16)/log(2) ~~ 1/(2*0.30103) = 1/0.60206 ~~ 1.66#