How do you evaluate #log_2 80#?

3 Answers
Binayaka C.
Jul 21, 2017

#log_2 80 = 6.32#

Explanation:

#log_2 80#

We know #log_a b = log_c b/log_c a# let #c=10 ;# then

#log_2 80 = log_10 80/log_10 2 ~~ 1.9031/0.3010~~ 6.32(2dp) # [Ans]

George C.
Jul 21, 2017

#log_2(80) ~~ 6.322#

Explanation:

Here's a way of calculating an approximation to #log_2 80# without the benefit of a calculator with #log# function...

First note that:

#2^6 = 64 < 80 < 128 = 2^7#

So:

#6 < log_2 80 < 7#

Now:

#log_2 80 = log_2 64 + log_2 (80/64) = 6 + log_2 (5/4)#

Next note that by squaring #5/4# we double its logarithm. So if the result is greater than #2# then #log_2 (5/4) > 1/2#

Let's try:

#(5/4)^2 = 25/16 < 2#

So:

#0 < log_2(5/4) < 1/2#

If we square again, we double again, so can test whether we need to add #1/4# to our estimated logarithm, etc.

#(25/16)^2 = 625/256 > 2#

So add #1/4# to our estimated logarithm and divide this "remainder" by #2# to get #625/512#.

Next:

#(625/512)^2 = 390625/262144 < 2#

So don't add #1/8#, but move on to:

#(390625/262144)^2 = 152587890625/68719476736 ~~ 2.22 > 2#

So do add #1/16# and we're quite close to the logarithm with #6+1/4+1/16 = 505/80 ~~ 6.3#

#color(white)()#
If we enlist the help of a calculator - say a four function one with #8# digits - then we might try the repeated doubling and dividing by #2# a bit further...

#80/64 = 1.25" "color(red)(6)#

#1.25^2 = 1.5625" "color(grey)(cancel(1/2))#

#1.5625^2 ~~ 2.4414603" "color(red)(1/4)#

#2.4414603/2 ~~ 1.2207032#

#1.2207032^2 ~~ 1.4901163" "color(grey)(cancel(1/8))#

#1.4901163^2 ~~ 2.2204466" "color(red)(1/16)#

#2.2204466/2 ~~ 1.1102233#

#1.1102233^2 ~~ 1.2325958" "color(grey)(cancel(1/32))#

#1.2325958^2 ~~ 1.5192924" "color(grey)(cancel(1/64))#

#1.5192924^2 ~~ 2.3082494" "color(red)(1/128)#

#2.3082494/2 ~~ 1.1541247#

#1.1541247^2 ~~ 1.3320038" "color(grey)(cancel(1/256))#

#1.3320038^2 ~~ 1.7742341" "color(grey)(cancel(1/512))#

#1.7742341^2 ~~ 3.1479066" "color(red)(1/1024)#

#3.1479066/2 ~~ 1.5739533#

#1.5739533^2 ~~ 2.4773290" "color(red)(1/2048)#

So:

#log_2(80) ~~ 6+1/4+1/16+1/128+1/1024+1/2048 = 12947/2048 ~~ 6.322#

George C.
Jul 21, 2017

#log_2(80) ~~ 6.321928#

Explanation:

Alternatively, if we know that #log_10(2) ~~ 0.30103# (which is a very good approximation), then we find:

#log_2(80) = log_2 8 + log_2 10#

#color(white)(log_2(80)) = log_2 2^3 + 1/(log_10 2)#

#color(white)(log_2(80)) ~~ 3 + 1/0.30103#

#color(white)(log_2(80)) ~~ 6.321928#

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