How do you calculate #Log_3 63#?

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Answer 1 · 2016-07-16T12:02:19

#3.771#

Some calculators can work out logs with any base, but we can work it out without that function.

By definition If #log_a b= c," then " a^c = b#

Let #log_3 63 = x" then " 3^x = 63#

#log 3^x = log 63#

#xlog3 = log63#

#x = (log 63)/(log 3) = 3.771#

This is the same result we would get using the change of base law.

#log_3 63 = (log_10 63)/(log_10 3)#

#=3.771#