If ln ⁡ x is defined via an integral and e defined from ln ⁡ x, how would you prove that ln ⁡ x is the inverse of e x ?This is a somewhat technically specific question about the relationship between ln ⁡ x and e x given one possible definition of ln ⁡ xSuppose that you define ln ⁡ x as ln ⁡ x = ∫ 1 x d t t ln ⁡ x = ∫ 1 x d t t We can use the connection between the integral definition of ln ⁡ x and the harmonic series to show that ln ⁡ x grows unboundedly. This function is monotonically increasing, so it should have an inverse mapping from its codomain ( R ) to its domain ( R + ). Let's call that function ln − 1 ⁡ xSince ln ⁡ x grows unboundedly and is zero when x is 1, we can define e as the unique value such that ln ⁡ e = 1So here's my question: given these starting assumptions, how would you prove that ln − 1 ⁡ x = e x (or, equivalently, that ln ⁡ x = log e ⁡ x? I'm having a lot of trouble even seeing how you'd get started proving these facts, since basically every calculus fact I know about e and ln ⁡ x presumes this result to be true.Thanks!

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If   ln  ⁡  x is defined via an integral and   e defined from   ln  ⁡  x, how would you prove that   ln  ⁡  x is the inverse of       e    x  ?This is a somewhat technically specific question about the relationship between   ln  ⁡  x and       e    x   given one possible definition of   ln  ⁡  xSuppose that you define   ln  ⁡  x as  ln  ⁡  x  =      ∫    1    x                      d        t            t        ln  ⁡  x  =      ∫    1    x                      d        t            t      We can use the connection between the integral definition of   ln  ⁡  x and the harmonic series to show that   ln  ⁡  x grows unboundedly. This function is monotonically increasing, so it should have an inverse mapping from its codomain (      R  ) to its domain (            R      +      ). Let&#039;s call that function       ln          −      1        ⁡  xSince   ln  ⁡  x grows unboundedly and is zero when   x is   1, we can define   e as the unique value such that   ln  ⁡  e  =  1So here&#039;s my question: given these starting assumptions, how would you prove that       ln          −      1        ⁡  x  =      e    x   (or, equivalently, that   ln  ⁡  x  =      log    e    ⁡  x? I&#039;m having a lot of trouble even seeing how you&#039;d get started proving these facts, since basically every calculus fact I know about   e and   ln  ⁡  x presumes this result to be true.Thanks!

1s1zubv · Accepted Answer

If   f  (  x  )  =  log  ⁡  x is defined as a primitive of       1    x   for which   f  (  1  )  =  0, then   f  (  a  b  )  =  f  (  a  )  +  f  (  b  ) holds for any couple of positive real numbers. This gives that the inverse function   g  (  x  ) is a       C    1   function, satisfies   g  (  0  )  =  1 and the functional equation:                    (1)                    g        (        a        )        g        (        b        )        =        g        (        a        +        b        )            for any couple of real numbers   a  ,  b.   (  1  ) and differentiability grants that       g    ′    =  g, hence   g  (  x  )  =      e    x

Banguizb · Accepted Answer

The advantage of defining   ln  ⁡  x as an integral is that it gives a simple way to define       e    x   in elementary calculus (without resorting to infinite series, taking limits of rational powers of   e, or other methods).Once   ln  ⁡  x has been defined in this way, you can establish that   y  =  ln  ⁡  x is an increasing function whose range is       R  , so it has an inverse function with domain       R   which we can       d    e    f    i    n    e   to be   y  =      e    x  With this definition, we have the identities  ln  ⁡  (      e    x    )  =  x for all   x and       e          ln      ⁡      x        =  x for   x  &amp;gt;  0, and   ln  ⁡  e  =  1    ⟹        e          1        =  e

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