Solving a logarithmic polynomialI want to solve this equation for x: 1 2 π x ( e z 2 x ) x = ϵIs there a closed form for it, or does it have to be solved numerically? I can turn it in to this: 2 x log ⁡ ( e z ) − ( 2 x + 1 ) log ⁡ ( 2 x ) = log ⁡ ( ϵ 2 π )Which is in a form similar to a + b log ⁡ x + c n + d n log ⁡ n = 0, which is a deceptively simple equation that I don't know how to solve analytically. It feels very similar to a polynomial, but instead of powers of x it's increasing O ( x )Failing a closed form answer, is there a useful closed form bound?

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Solving a logarithmic polynomialI want to solve this equation for   x:            1              2        π        x                            (                              e            z                                2            x                          )            x        =  ϵIs there a closed form for it, or does it have to be solved numerically? I can turn it in to this:  2  x  log  ⁡  (  e  z  )  −  (  2  x  +  1  )  log  ⁡  (  2  x  )  =  log  ⁡  (      ϵ    2    π  )Which is in a form similar to   a  +  b     log  ⁡  x  +  c     n  +  d     n  log  ⁡  n  =  0, which is a deceptively simple equation that I don&#039;t know how to solve analytically. It feels very similar to a polynomial, but instead of powers of   x it&#039;s increasing       O    (  x  )Failing a closed form answer, is there a useful closed form bound?

Haggar72 · Accepted Answer

AFAIK there is no closed form solution to the equation.Assuming   e  z  ,  x  ,  ϵ  &amp;gt;  0, as   z  →  ∞ you have an asymptotic solution  x  =                    e        z            2        −                    ln        ⁡        (        e                  ϵ          2                π        )            2        −                    ln        ⁡        (        z        )            2        +  O  (  1      /    z  )

Yahir Tucker · Accepted Answer

Imagine if you had       x          x      +      1        =  5. If you can get your equation into that type of form you will be in a better position.Let&#039;s deal with this hypothetical equation because it is simpler and easier to manipulate. And if we can solve this equation, it will be easy to go back and solve your equation, just some of the constants will be different.First thing, we have       log    x    ⁡  5  =  x  +  1We need to get that base   x out of there. What if we went   x  =      5                  1                  x          +          1                     then take the log of that?      log    5    ⁡  x  =      1          x      +      1            (  x  +  1  )      log    5    ⁡  x  =  1Do you know the general formula for multiplying two log expressions?      log                  x        a              ⁡      x    y    ×      log                  x        b              ⁡      x    y    =      log                  x                  a          b                      /                    y                      ⁡      x    y  Choose a convenient substitute   x that will work, such that       x    y   is equal to your   x, and the   x makes for a convenient base.In these types of equations, eventually you are going to move the   x variable into the exponent, leaving behind constantsI mean you will have constant   A raised to the exponent of the log of   x multiplied by constant   B raised to a different base log of   x. Making the constant   B into constant   A is easy if you just move it into the exponent. Once the two constants are the same, the exponents can be added together over the same constant.Adding together two logs of   x with different log bases is not that difficult. Make the first log term equal to variable   a, and the other log term equal to   b. Use some substitution.

Solving a logarithmic polynomial I want to solve this equation for x : <mrow class="MJX-Te

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