Solving equations of type x 1 / n = log n ⁡ xFirst, I'm a new person on this site, so please correct me if I'm asking the question in a wrong way.I thought I'm not a big fan of maths, but recently I've stumbled upon one interesting fact, which I'm trying to find an explanation for. I've noticed that graphs of functions y = x 1 / n and y = log n ⁡ x , where n is given and equal for both functions, always have 2 intersection points. This means, equation x 1 / n = log n ⁡ x must have 2 solutions, at least it's what I see from the graphs.I've tried to solve this equation analytically for some given n, like 4, but my skills are very rusty, and I cannot come up with anything. So I'm here for help, and my question(-s) are:are these 2 functions always have 2 intersection points?if yes, why, if not, when not?how to solve equations like x 1 / n = log n ⁡ x analytically?

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Solving equations of type       x          1              /            n        =      log          n        ⁡  xFirst, I&#039;m a new person on this site, so please correct me if I&#039;m asking the question in a wrong way.I thought I&#039;m not a big fan of maths, but recently I&#039;ve stumbled upon one interesting fact, which I&#039;m trying to find an explanation for. I&#039;ve noticed that graphs of functions   y  =      x          1              /            n       and   y  =      log          n        ⁡  x , where   n is given and equal for both functions, always have   2 intersection points. This means, equation       x          1              /            n        =      log          n        ⁡  x must have   2 solutions, at least it&#039;s what I see from the graphs.I&#039;ve tried to solve this equation analytically for some given   n, like   4, but my skills are very rusty, and I cannot come up with anything. So I&#039;m here for help, and my question(-s) are:are these   2 functions always have   2 intersection points?if yes, why, if not, when not?how to solve equations like       x          1              /            n        =      log          n        ⁡  x analytically?

benedictazk · Accepted Answer

x          n                                    =                  log          n                ⁡        x                                          x                            =                  t          n                                   }          ⟺    t  =      n          ln      ⁡      n        ⋅  ln  ⁡  t    ;    t  =      e    u      ⟺        e    u    =      n          ln      ⁡      n        ⋅  u    ⟺    (  −  u  )  ⋅      e          −      u        =  −            ln      ⁡      n        n      ⟺    u  =  −  W            (        −            ln      ⁡      n        n              )          ⟺    x  =      t    n    =  (      e    u        )    n    =      e          n      u        x  =  exp  ⁡            (        −  n  ⋅  W            (        −            ln      ⁡      n        n              )                  )      where W is the Lambert W function.

abbracciopj · Accepted Answer

They always have two intersection points. Let  f  (  x  )  =      x          1              /            n        −      log    n    ⁡  x  .Then      lim          x      →              0        +              f  (  x  )  =      lim          x      →      +      ∞        f  (  x  )  =  +  ∞  .Also,      f    ′    (  x  )  =      1    n          x          1              /            n      −      1        −      1          x            log      ⁡      n        .It is easy to see that       f    ′   has a single zero, which is necessarily a minimum (one can check that       f    ′   has the appropriate signs at both sides of this point, which is      x    m    =            n      n              (      log      ⁡      n              )        n              .We have  f  (      x    m    )  =      n          log      ⁡      n        −      n          log      ⁡      n          (  log  ⁡  n  −  log  ⁡  log  ⁡  n  )  &amp;lt;  0  ,so the minimum is achieved below the   x-axis. This shows that   f intersects the   x-axis twice.As for an analytic solution, I don&#039;t think that&#039;s possible.

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