We can expand #log_9(sqrt(sqrt(sqrt(3))))# to #log_9(((3^(1/2))^(1/2))^(1/2))# (since the square root is equivalent to taking the power of one half).

Now, #(a^b)^c=a^(bc)#. Using this, we simplify #log_9(((3^(1/2))^(1/2))^(1/2))# to #log_9(3^(1/2*1/2*1/2))=log_9(3^(1/8))#.

From here, we just need to remember the logarithm identity #log(a^b)=blog(a)#. Using this identity, we simplify #log_9(3^(1/8))# to #log_9(3)/8#.

But what is #log_9(3)#? Remember that taking the #log_b(a)# is finding a number #n# so that #b^n=a#. We know that #sqrt(9)=9^(1/2)=3#. So, #log_9(3)=1/2#.

Then, #log_9(3)/8=(1/2)/8=1/16#.