Q:  If Cepheids have an absolute magnitude which is up to 6 magnitudes
           brighter than RR Lyr, how much farther should we be able
           to see them?


         Let's define

                       d1  =  max distance to RR Lyr
                       d2  =  max distance to Cepheid

                       I1  =  intensity of RR Lyr at max distance
                       I2  =  intensity of Cepheid at max distance

         From the inverse square law, we know

                                            2
                             I1       ( d2 )
                            ----  =  -------
                             I2       ( d1 )


         Based on their magnitudes,

                        m1  =  apparent magnitude of RR Lyr at max distance
                        m2  =  apparent magnitude of Cepheid at max distance

    
                                                  I1
                      m1 - m2   =   -2.5*log10 ( --- )
                                                  I2

  
          Substitute for I1/I2
                                                      2
                                               (  d2 )
                      m1 - m2   =   -2.5*log10 ( --- )
                                               (  d1 )


                                                      
                                               (  d2 )
                      m1 - m2   =   -5.0*log10 ( --- )
                                               (  d1 )

                                                      

           Now solve for the ratio of distances



                        d2          -0.2*(m1 - m2)
                       ----   =   10
                        d1


           Let's use the fact that Cepheids may be up to 6 magnitudes
           brighter, so (m1 - m2) = -6.


                        d2
                       ----  =  16
                        d1


           So we can observe and measure Cepheids perhaps 16 times
           more distant than RR Lyr stars.