I'm having a set of points in the form:

${(\frac A{255})}^{\frac 1x}=\frac B{255}$

I need to Find $x$ in the Equation.Where $A$ & $B$ are set of constants ranging from $0$ to $255$. Please provide me some methods to solve this equation with explanation. Assume I'm having very little knowledge in this math involved.

  • $\begingroup$ What about taking logarithms of both sides ? This should be the first step. $\endgroup$ – Claude Leibovici Apr 9 '15 at 12:21
  • $\begingroup$ Can you please elaborate it? $\endgroup$ – Balaji R Apr 9 '15 at 12:26

If I got right what are you asking I would take the log on both sides which leads me to:

$\frac 1x=\log_{\frac A{255}}(\frac B{255})$

And then (taking reciprocals and using logs properties):

$x=\log_{\frac B{255}}(\frac A{255})$

If you need to computate this with a high degree of accuracy you can first say that:

1) $A$ and $B$ must be different from $0$ for the existence of the log;

2) $B$ must be different from $255$ for the existence of the log's base;

3) If the base exists and $A=255$ then $x$ would be $0$ but it can't be for the existence of the exponent.

Now you can convert the log in base $10$ or base $e$ (natural log) to make calculus faster for your calculator.

  • $\begingroup$ Why not to use $\log_e$ or $\log_{10}$ ? $\endgroup$ – Claude Leibovici Apr 9 '15 at 12:38
  • $\begingroup$ Using the base $\frac A{255}$ I can get straight to the result whitout changing base. $\endgroup$ – Renato Faraone Apr 9 '15 at 12:39
  • $\begingroup$ You are perfectly correct from a mathematical point of view. But how would you compute $x$ for given $A,B$ ? $\endgroup$ – Claude Leibovici Apr 9 '15 at 12:48
  • $\begingroup$ So for log 10 ,Result will be x = log (A/255) / log (B/255). is this right? $\endgroup$ – Balaji R Apr 9 '15 at 13:14
  • 1
    $\begingroup$ Edited with the hope now things are more precise ;) $\endgroup$ – Renato Faraone Apr 9 '15 at 14:51

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