How do you evaluate #log_81 3#?

2 Answers
Nov 2, 2016

#log_81 3 = 1/4#

Explanation:

#log_81 3 =x#

Rewrite as an exponential. Remember, the answer to a log is the exponent. In this case #x# is the exponent, and #81# is the base.

#81^x =3#

Find a common base for both sides, which is #3#.
#81=3^4#, so by substitution...

#(3^4)^x=3#

Use the exponent rule #(x^a)^b=x^(ab)#

#3^(4x)=3^1#

#4x=1#

#x=1/4#

Nov 2, 2016

#log_81(3)=1/4#

Explanation:

Since #81# is much larger than #3#, our answer will be a decimal, so let's think of this problem in the opposite sense: #log_3(81)#. #3^4 = 81#, so #log_3(81) = 4#.

Using the law of exponents, we know that if #a^m = n#, then #n^(1/m) = a#. So using this rule, we know that #81^(1/4) = 3#, so #log_81(3)=1/4#