Evaluate. or calculate the value

-- The difference of 2 logs is the log of the quotient of their arguments.

             log(11) - log(6) = log(11/6)

-- 1/3 of the log of something is the log of its cube root.
 
                     1/3 log(8) = log(∛8) = log(2)
and
                     1/3 log(729) = log(∛729) = log(9)

-- If a bunch of logs all have the same base, then their sum
is the log of the product of the arguments.  So ...

         log(11) - log(6) + 1/3 log(8) + 1/3 log(729)  =

         log(11/6 times 2 times 9)  =

         log( 11*18 / 6 )  =  log(33)

                                      log(33) = about 1.519 (rounded)



The other way:

log(11) =                  1.0414
-log(6) =                  -0.7782
log(8) = 0.9031
       1/3(0.9031) =     0.3010
log(729) = 2.8627
           1/3(2.8627) = 0.9542
                             
Adum up:                 1.5184

(Note:  Everything is rounded.)

Hope i can you!