How do you write #9^(3/2)=27# in logarithmic form?

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Answer 1 · 2017-01-08T20:24:30

#log_9(27)=3/2#

Compare to a known condition

Suppose we had #" "log_10(x)=y#

This is the same as #" "10^y=x" " ul(vec("compare to"))" " 9^(3/2)=27#

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So in the case of the question:

Instead of log to base 10 #-> log_10# we have log to base 9#-> log_9#

For the index of #y# we have #3/2#

For the answer of #x# we have 27

Answer 2 · 2017-01-08T21:32:58

#log_9 27 = 3/2#

#log_ab= c# can be changed to #a^c = b#

For the logarithmic function #9^(3/2) = 27#

#a = 9#
#b = 27#
#c = 3/2#

This can become

#log_9 27 = 3/2#