How do you evaluate #log_216 6#?

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Answer 1 · 2016-08-29T20:55:06

#1/3#

#6^2 = 36#

#6^3 = 36 * 6 = 180+36 = 216#

So #216^(1/3) = 6#

and #log_216 6 = 1/3#

Answer 2 · 2016-08-29T20:55:24

#log_216 6 = 1/3#

By the #log# rule #log_a(n) = (logn)/(loga)#:

#log_216(6) = (log6)/(log216) = (log(6^1))/(log(6^3))#

Now use the rule #log(a^n) = nloga# and simplify:

#=(1log6)/(3log6)#

#=1/3#