Question #03c7e

1 Answer
Dec 21, 2017

#approx 4.90 #

Explanation:

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