How do you evaluate #log_9 x=3/2#?

1 Answer
Oct 27, 2016

#x=27#

Explanation:

Remember that in general #log_b a = c# means #b^c=a#
(You really need to memorize this; it's a basic definition.)

Therefore #log_9 x=3/2#
means
#color(white)("XXX")9^(3/2)=x#

#rarrcolor(white)("XX")x=9^(3/2)#

#color(white)("XXXX")=(9^(1/2))^3#

#color(white)("XXXX")=(+3)^3# (the negative root can be ignored since log arguments must be positive)

#color(white)("XXXX")=27#