Q:  If Vega is 10 times brighter than Mintaka, what is the
            magnitude of Mintaka?


         First, recall that Vega has a magnitude of zero (or close to it).
         Therefore, if we can find the difference in magnitudes between
         Vega and Mintaka, that will be the same as the magnitude of Mintaka.


         The ordinary, long way to perform this calculation goes like this:


                                                  intensity(Mintaka)
           m(Mintaka) - m(Vega)  =   -2.5 log10 ( ------------------- )
                                                     intensity(Vega)


                                                    1
                                 =   -2.5 log10 ( ----- )
                                                   10


                                 =   -2.5 ( -1 )  =  2.5


          Therefore,

                 magnitude of Mintaka    =   magnitude of Vega  +  2.5

                                         =       0  +  2.5
 
                                         =   2.5



          The short cut is to realize that a ratio of 10 in intensity
          is "halfway", logarithmically, to a ratio of 100; and therefore
          the difference in magnitudes will be halfway to the difference
          corresponding to a ratio of 100.

                   ratio of 100      --->   difference in magnitude of 5

            10 is "halfway" to 100   --->   difference is half of 5 = 2.5