Summing reciprocal logs of different basesI recently took a math test that had the following problem: 1 log 2 ⁡ 50 ! + 1 log 3 ⁡ 50 ! + 1 log 4 ⁡ 50 ! + ⋯ + 1 log 50 ⁡ 50 ! The sum is equal to 1. I understand that the logs can be broken down into (first fraction shown) 1 log 2 ⁡ 1 + log 2 ⁡ 2 + log 2 ⁡ 3 + ⋯ + log 2 ⁡ 50 How do the fractions with such irrational values become 1? Is there a formula or does one simply need to combine fractions and use the basic properties of logs?

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Summing reciprocal logs of different basesI recently took a math test that had the following problem:      1                  log                  2                    ⁡      50      !        +      1                  log                  3                    ⁡      50      !        +      1                  log                  4                    ⁡      50      !        +  ⋯  +      1                  log                  50                    ⁡      50      !      The sum is equal to 1. I understand that the logs can be broken down into (first fraction shown)      1                  log                  2                    ⁡      1      +              log                  2                    ⁡      2      +              log                  2                    ⁡      3      +      ⋯      +              log                  2                    ⁡      50      How do the fractions with such irrational values become   1? Is there a formula or does one simply need to combine fractions and use the basic properties of logs?

last99erib · Accepted Answer

1                  log                  2                    ⁡      50      !        +      1                  log                  3                    ⁡      50      !        +      1                  log                  4                    ⁡      50      !        +  ⋯  +      1                  log                  50                    ⁡      50      !        =      log          50      !        ⁡  2  +      log          50      !        ⁡  3  +      log          50      !        ⁡  4  +  .  .  .  +      log          50      !        ⁡  50  =      log          50      !        ⁡  (  2  ⋅  3  ⋅  4  ⋅  .  .  .50  )  =      log          50      !        ⁡  (  50  !  )  =  1

deceptie3j · Accepted Answer

This is a general result. We can write for any   N                              ∑                      n            =            2                    N                          1                                    ∑                              m                =                2                            N                                      log              n                        ⁡            (            m            )                                              =                  ∑                      n            =            2                    N                          1                                    ∑                              m                =                2                            N                                                                        log                  b                                ⁡                (                m                )                                                              log                  b                                ⁡                (                n                )                                                                                                    =                                            ∑                              n                =                2                            N                                      log              b                        ⁡            (            n            )                                              ∑                              m                =                2                            N                                      log              b                        ⁡            (            m            )                                                                          =        1            where we used       log    n    ⁡  (  m  )  =                    log        b            ⁡      (      n      )                      log        b            ⁡      (      m      )      . And we are done!

Summing reciprocal logs of different bases I recently took a math test that had the following probl

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