4 log 2 ⁡ ( n ) = n How to Solve for n?As the title suggests, my log skills are pretty lacking. Need to learn how to get from 4 log 2 ⁡ ( n ) = n to n = 16 ( log base 2). I've searched Google and it seems I am missing some core concept here. I really appreciate the help or guidance.

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4      log    2    ⁡  (  n  )  =  n How to Solve for   n?As the title suggests, my log skills are pretty lacking. Need to learn how to get from   4      log    2    ⁡  (  n  )  =  n  to   n  =  16 (  log base   2). I&#039;ve searched Google and it seems I am missing some core concept here. I really appreciate the help or guidance.

Colin Moran · Accepted Answer

You need to use the Lambert W-function, which can&#039;t be solved with elementary functions (but can be expressed as an infinite sum using the Lagrange inversion forumula)However just by trying random powers of 2 you can get an integer solution of   n  =  16, probably the easiest way.

veirarer · Accepted Answer

4      log    2    ⁡  n  =  n    ⟺        2          4              log        2            ⁡      n        =      2    n      ⟺        n    4    =      2    n      ⟺    n  =      2    k   , since   n  ∈      N      ⟺        2          4      k        =      2                  2        k              →  →  4  k  =      2    k      ⟺    k  =      2          k      −      2      . Can you take it from here ? :-)  4      log    2    ⁡  n  =  n    ⟺    4                    ln        ⁡        n                    ln        ⁡        2              =  n    ⟺                      ln        ⁡        n            n        =                                                        ln              ⁡              2                        4                          ⏟              constant    .   Now draw the graphic of                     ln        ⁡        x            x       , and tell me what you notice. :-)

4 log 2 </msub> ⁡ ( n ) = n How to Solve for n ? A

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