How do you evaluate the expression #log_9 243#?

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Answer 1 · 2016-11-04T13:27:48

#log_(9)243=5/2#

#log_ab=c=>a^c=b#

#y=log_(9)243=>9^y=243#

recognising that #243# is a power of #3#

# 3^5=243#

and #9^(1/2)=3#

we have# (9^(1/2))^5=243#

#:.9^(5/2)=243#

#y=5/2#