How do you evaluate #log_27 3#?

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Answer 1 · 2016-08-28T21:22:47

#log_27 3 = 1/3#

Try to read log expressions in this form as a question...

#3 " is which power of " 27?#

OR: #"What index of 27 will give "3?#

You should recognise that 3 is the cube root of 27

In index form this is: #27^x = 3#

#root3 (27) = 3 " or 27^(1/3) = 3#.

#log_27 3 = 1/3#

In the same way try to answer these by inspection:

#log_7 49, " "log_5 125 , " " log_36 6 " " log_32 2#

The answers in reverse order are:
#1/5," " 1/2" " 3, " " 2#