0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.