Why does the method to find out log and cube roots work?To find cube roots of any number with a simple calculator, the following method was given to us by our teacher, which is accurate to atleast one-tenths.1)Take the number X, whose cube root needs to be found out, and take its square root 13 times (or 10 times) i.e. . . . . X 2)next, subtract 1, divide by 3 (for cube root, and any number n for nth root), add 1.3) Then square the resultant number (say c) 13times (or 10 times if you had taken out root 10 times) i.e. c 2 2 . . . .2 = c 2 13 . This yields the answer.I am not sure whether taking the square root and the squares is limited to 10/13 times, but what I know is this method does yield answers accurate to atleast one-tenths.For finding the log, the method is similar:-1)Take 13 times square root of the number, subtract 1, and multiply by 3558. This yield s the answer.

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Why does the method to find out log and cube roots work?To find cube roots of any number with a simple calculator, the following method was given to us by our teacher, which is accurate to atleast one-tenths.1)Take the number   X, whose cube root needs to be found out, and take its square root 13 times (or 10 times) i.e.                               .          .          .          .          X                    2)next, subtract 1, divide by 3 (for cube root, and any number n for nth root), add 1.3) Then square the resultant number (say c) 13times (or 10 times if you had taken out root 10 times) i.e.       c                  2                              2                          .              .              .              .2                                            =      c                  2                  13                    . This yields the answer.I am not sure whether taking the square root and the squares is limited to 10/13 times, but what I know is this method does yield answers accurate to atleast one-tenths.For finding the log, the method is similar:-1)Take 13 times square root of the number, subtract 1, and multiply by   3558. This yield s the answer.

Paxton James · Accepted Answer

Let&#039;s use these classical formulae :      e    x    =      lim          n      →      ∞                  (      1      +              x        n            )        n    ln    x  =      lim          n      →      ∞        n      (          x              1                  /                n              −    1    )  to get (replacing the limit by a large enough value of   n:  N  =      2          13      ) :                              x          3                =                  e                                    (              ln              ⁡              x                              /                            3              )                                                          ≈                              (            1            +                                          ln                ⁡                x                                  /                                3                            N                        )                    N                                                  ≈                              (            1            +                                          N                                  (                                      x                                          1                                              /                                            N                                                        −                  1                  )                                                            3                                N                                      )                    N                                                  ≈                              (            1            +                                          (                                  x                                      1                                          /                                        N                                                  −                1                )                            3                        )                    N                    Concerning the decimal logarithm we have :      log          10          x  =            ln            x              ln            10        ≈      N          ln            10            (          x              1                  /                N              −    1    )  For   N  =      2          13        we may (as indicated by peterwhy) approximate the fraction with      N          ln            10        =            2              13                    ln            10        ≈  0.4343  ×  8192  ≈  3558Hoping this clarified things,

Why does the method to find out log and cube roots work? To find cube roots of any number with a si

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