The first thing we can do is use our log laws:
#logalpha + log beta = log alpha beta #
So for this problem:
#log_4 (16*55.715) = log_4 16 + log_4 55.715 #
We know #log_4 16 = 2 #
Now we can start to approximate #log_4 55.715: #
#=> log_4 16 < log_4 55.715 < log_4 64 #
#=> 2 < log_4 55.715 < 3 #
But we can go even further with our approximates...
#=> log_4 32 < log_4 55.715 < log_4 64 #
#=> 5/2 < log_4 55.715 < 3 #
We can then yield...
#=> 5/2 + log_4 16 < log_4 (16*55.715) < 3 + log_4 16 #
#=> 9/2 < log_4 (16*55.715) < 5 #
Hence taking the arithmetic mean of #9/2 and 5 #:
We can say #log_4 ( 16*55.715) approx (9/2 + 5 )/2 approx 19/4 #
This is a half decent approximation, but if we were wanting to be exact, typing into a calculator:
#log_4 (16*55.715 ) approx 4.899996944#
#color(red)(approx 4.90 # to 2 d.p
So our approximate of #19/4# has a #3.06% # error
