How to show log ⁡ a ≤ n ( a n − 1 ) ≤ a n log ⁡ aLet b = a n How to show: log ⁡ a ≤ n ( a n − 1 ) ≤ a n log ⁡ a?Thank you ;)

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How to show   log  ⁡  a  ≤  n  (      a    n    −  1  )  ≤      a    n    log  ⁡  aLet   b  =      a    n  How to show:   log  ⁡  a  ≤  n  (      a    n    −  1  )  ≤      a    n    log  ⁡  a?Thank you ;)

tiltat9h · Accepted Answer

Noting   a  =      b    n  , this is equivalent to  n  log  ⁡  b  ≤  n  (  b  −  1  )  ≤  n  b  log  ⁡  b  ,i.e. justTo get this, employ the mean value theorem separately for   b  ≥  1 and   0  &amp;lt;  b  &amp;lt;  1 (I assume the condition   b  &amp;gt;  0 as   log isn&#039;t defined for non-positive numbers). Notice first it does work for   b  =  1, as all 3 quantities are   0. If   b  &amp;gt;  1, applying the mean value theorem to   x  ↦  log  ⁡  x gives us that there exists   c  ∈  [  1  ,  b  ] such that  log  ⁡  b  −  log  ⁡  1  =      1    c    (  b  −  1  )  ,then noting   log  ⁡  1  =  0, and using the facts   c  ∈  [  1  ,  b  ] and   b  −  1 are positive gives us      1    b    (  b  −  1  )  ≤      1    c    (  b  −  1  )  ≤  b  −  1gives us the result we want after a little bit of re-arranging. For   b  &amp;lt;  1, the working is pretty much the same, except we get the last inequality from the fact   c  ∈  [  b  ,  1  ] and   b  −  1 is negative instead.

How to show log a<= n(root[n](a)-1) <= sqrt[n](a) log a

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