How do you evaluate #log_216 6#?

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Answer 1 · 2016-09-09T03:07:34

#log_(216)6=1/3#

A couple of things to know/remember to answer this:

#log_(216)6# is asking how many times 216 needs to be multiplied by itself in order to arrive at 6. Clearly the answer is less than 1!

#216^(1/2)# is the same as #sqrt216#. #216^(1/3)# is the same as #root(3)216#. And so on.

Ok... what is #log_(216)6#?

One way to do this is to work from the 6 to the 216 - it turns out that:

#6^3=216#

Which means that

#216^(1/3)=6#

and so:

#log_(216)6=1/3#