Show that ln ⁡ ( x ) ≤ x − 1I'm not really sure how to show this, it's obvious if we draw a graph of it but that won't suffice here. Could we somehow use the fact that e x is the inverse? I mean, if e x − 1 ≥ x then would the statement be proved?

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Show that   ln  ⁡  (  x  )  ≤  x  −  1I&#039;m not really sure how to show this, it&#039;s obvious if we draw a graph of it but that won&#039;t suffice here. Could we somehow use the fact that       e    x   is the inverse? I mean, if       e          x      −      1        ≥  x then would the statement be proved?

Bobby Mcconnell · Accepted Answer

Yes, one can use                    (1)                              e          x                ≥        1        +        x        ,            which holds for all   x  ∈      R   (and can be dubbed the most useful inequality involving the exponential function). This again can be shown in several ways. If you defined       e    x   as limit       lim          n      →      ∞                  (      1      +              x        n            )        n  , then   (  1  ) follows from Bernoullis inequality:   (  1  +  t      )    n    &amp;gt;  1  +  n  t if   t  &amp;gt;  −  1 and   n  &amp;gt;  0.To show that   ln  ⁡  (  x  )  ≤  x  −  1 for all   x  &amp;gt;  0, just substitute   ln  ⁡  x for x in (1).

Hunter Shah · Accepted Answer

Define for     x  &amp;gt;  0    f  (  x  )  =  ln  ⁡  x  −  x  +  1    ⟹        f    ′    (  x  )  =      1    x    −  1  =  0    ⟺    x  =  1and since         f    ″    (  x  )  =  −            1              x        2              &amp;lt;  0    ∀  x  &amp;gt;  0   , we get a maximal point.But also      lim          x      →      0      +        f  (  x  )  =  −  ∞  =      lim          x      →      ∞        f  (  x  )Thus, the above is a global maximal point and  ∀    x  &amp;gt;  0    ,        f  (  x  )  ≤  f  (  1  )  =  0

Show that ln(x)<=x−1

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