How do you evaluate #Log_sqrt3 243 #?

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Answer 1 · 2016-03-13T13:56:53

#10#

We should try to write #243# as a power of #3#.

An example of the method we should attempt can be shown more easily in evaluating #log_2 8#:

#log_2 8=log_2 2^3=3log_2 2=3#

The first step we should take for #log_sqrt3 243# is to recognize that #243=3^5#.

#log_sqrt3 243=log_sqrt3 3^5#

But we still haven't addressed the issue that we want a base of #sqrt3#, not #3#.

We should use the fact that #(sqrt3)^2=3#, like so:

#log_sqrt3 3^5=log_sqrt3 ((sqrt3)^2)^5#

Using the rule that #(x^a)^b=x^(ab)#, we multiply #2# and #5# to see that

#log_sqrt3 ((sqrt3)^2)^5=log_sqrt3 (sqrt3)^10#

Now, simplify as was done earlier.

#log_sqrt3 (sqrt3)^10=10log_sqrt3sqrt3=10#