If x x y y z z = c, show that at x = y = z , z x y = − ( x log ⁡ x ) − 1 I applied logarithmic functions on both sides, ultimately it led to a satisfying result where most things canceled out, but unfortunately, it wasn't the right answer.

Question

If       x    x        y    y        z    z    =  c, show that at   x  =  y  =  z  ,      z          x      y        =  −  (  x  log  ⁡  x      )          −      1      I applied logarithmic functions on both sides, ultimately it led to a satisfying result where most things canceled out, but unfortunately, it wasn&#039;t the right answer.

Dillan Foley · Accepted Answer

Taking logarithms of both sides and differentiating wrt x gives  log  ⁡  x  +  1  +      z    x    log  ⁡  z  +      z    x    =  0.Now differentiate wrt y to get      z          x      y        log  ⁡  z  +      z    x        1    z        z    y    +      z          x      y        =  0.From the first equation you can solve for zx and substitute in the second equation. Finally, differentiating the original equation (after taking logs of course) now wrt y, we get  log  ⁡  y  +  1  +      z    y    log  ⁡  z  +      z    y    =  0  ,from which again you may solve for zy and substitute in the equation for zxy. Then finally you may express zxy explicitly in terms of x,y,z. Finally, set x=y=z, and you&#039;d have been done.Or, more expeditiously, note that when x=y=z, we have       z    x    =      z    y    =  −  1. This gives      z          x      y        =  −      1    x        1          1      +      log      ⁡      x        ,so that your result is closer to the truth than the book&#039;s. I suspect there was a mistyping by the book makers.

If x^x y^y z^z=c, show that at x=y=z,z_(xy)=-(x log x)^(-1)

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