How do you evaluate #log_27 9#?

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Answer 1 · 2016-08-07T18:16:41

#x = 2/3#

In log form the question being asked is ...

"To what power must 27 be raised to give 9?#

You should recognise 27 as #3^3# and 9 as #3^2#

#3 = root3(27) = 27^(1/3)#

#9 = 3^2 = (root3(27))^2 = 27^((1/3)^2) = 27^(2/3)#

OR

#log_27 9 = x rArr " " 27^x = 9#

#3^(3x) = 3^2#

Equating exponents: #3x = 2#

#x = 2/3#