How do you evaluate #log_27 9#?

1 Answer
Aug 7, 2016

#x = 2/3#

Explanation:

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#