How do you evaluate #log_3 64#?

2 Answers
Gió
Dec 26, 2016

I found #3.78557#

Explanation:

I would try to change base and use a pocket calculator. The natural log, #ln#, can normally be evaluated using a pocket calculator so I 'll try to use #ln#:
#log _3(64)=(ln(64))/(ln(3))=3.78557#

Remember that to change base to a new base #c# you do:
#log_b(a)=(log_c(a))/(log_c(b))#

George C.
Dec 26, 2016

#log_3 64 = (6 log 2)/(log 3) ~~ 3.7855786#

Explanation:

Suppose you know the following approximations:

#log_10 2 ~~ 0.30103#

#log_10 3 ~~ 0.47712125#

The change of base formula tells us that:

#log_a b = (log_c b)/(log_c a)#

for any #a, b, c > 0# with #a, c != 1#

So we find:

#log_3 64 = (log_10 64)/(log_10 3) = (log_10 2^6)/(log_10 3) = (6 log_10 2)/(log_10 3) ~~ (6*0.30103)/(0.47712125) ~~ 3.7855786#

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