How do I find #log_27 9#?

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Answer 1 · 2015-03-15T00:05:06

The answer is #2/3#.

Method 1
#log_27(9)# can be interpreted as "#27# to what power is equal to #9#.

Since #27^(2/3) = 3^2 = 9, log_27(9) = 2/3#.

Method 2
If it is difficult to see directly, another technique is to change the base:

#log_27(9)# can be rewritten as #(log_3(9))/log_3(27) = 2/3#

Answer 2 · 2015-03-15T03:21:34

As has already been pointed out, #log_27(9)# is the exponent needed on 27 to get 9. It is the solution to #27^x=9#.

#27^x=9#

Because #27=3^3# and #9=3^2#, we must want

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

But #(3^3)^x=3^(3x)#, so we also want:

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

Now, it is a property of exponents the: if #3# to this power equals #3# to that power, then this power must be equal to that power.

So we need #3x=2#, which means we need #x=2/3.

Check: #27^(2/3)=root(3)27^2=3^2=9#